Class 11 Engineering Maths - Binomial Theorem

Find the middle terms in the expansion of $(5x-7y)^7$ .

Since n is an odd number, the expansion contains two middle terms.

$\displaystyle\left(\frac{7+1}{2}\right)^{th}$ and

$\displaystyle\left(\frac{7+3}{2}\right)^{th}$ terms are the two middle terms in the expansion of

$(5x-7y)^7$ .

$T_4=T_{3+1}=(-1)^3\:^7C_3(5x)^{7-3}(7y)^3=-^7C_3(5x)^4(7y)^3$

$T_5=T_{4+1}=(-1)^4\:^7C_4(5x)^{7-4}(7y)^4=-^7C_4(5x)^3(7y)^4$

Find the 28th term of $(5x+8y)^{30}$ .

$T_{27+1}={}^{30}C_{27}(5x)^{30-27}(8y)^{27}=\dfrac {30!}{3!27!}(5x)^3.(8y)^{27}$

Show that the expansion of $\left( x^2 + \dfrac{1}{x} \right)^{12}$ does not contain any term involving $x^{-1}$ .

General term is

$T_{r} = ^{n}C_{r}\,x^{r}\,a^{n-r}$

$(x+a)r_{n}(r\epsilon z)$

$\therefore T_{r} = ^{12}C_{r}(x^{2})^{r}\left ( \dfrac{1}{x} \right )^{12-r}$

$\Rightarrow T_{r} = ^{12}C_{r}\,x^{3r-12}$

for

$x^{-1}$ term

$3r-12 = -1$

$\Rightarrow \boxed{r = \dfrac{11}{3}}$ not possible

1115303_529248_ans_f0c69f76e77a466bb816d0e711ae9d55.jpg

Find the middle term in the expansion ${ \left( \cfrac { x }{ a } +\cfrac { y }{ b } \right) }^{ 6 }$ .

Given expression,

${ \left( \dfrac { x }{ a } +\dfrac { y }{ b } \right) }^{ 6 }$

No of terms will be

$6+1=7$ terms

$\therefore$ middle term will be

$4^{th}$ term

$t_4={ }^6C_3{ \left( \dfrac { x }{ a } \right) }^{ 3 }{ \left( \dfrac { y }{ b } \right) }^{ 6-3 }$

$t_4=20\times\dfrac { { x }^{ 3 }{ y }^{ 3 } }{ { a }^{ 3 }{ b }^{ 3 } }$ is the middle term.

Write down and simplify:
The 4th term of ${ \left( x-5 \right) }^{ 13 }$

We know that the

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ the fourth term in expansion of

$(x-5)^{13}$ is

$^{13}C_3 x^{10}(-5)^3 =-35750x^{10}$

Find the $13^{th}$ term in the expansion of ${ \left( 9x-\cfrac { 1 }{ 3\sqrt { x } } \right) }^{18 }$ .

in a binomial expansion

$(a+b)^n$ the

$r+1th$ term is

${ ^n{ C }_{ r } }{ a }^{ n-r }{ b }^{ r }$

here

$n=18,r=12,a=9x,b$ =

$\dfrac{1}{3x^{1/2}}$

therefore by substituing we get

${ ^{18} }{ { C }_{ 12 } }{ (9x) }^{ 6 }{ (\frac { -1 }{ 3\sqrt { x } } ) }^{ 12 }={ ^{18} }{ { C }_{ 12 } }=18564$

The sum of the coefficients in the first three terms of the expansion of $\displaystyle\, \left ( x^2 - \frac{2}{x} \right )^m$ is equal to $97$ . Find the term of the expansion containing $x^4$ .

Sum of first three terms be

$^{m}C_{0}-2 ^{m}C_{1}+4^{m}C_{2}=97$

$\implies 1-2{m}+2(m)(m-1)=97\implies 2{m^2}-4{m}-96=0\implies m^{2}-2{m}-48=0\implies m=8,-6$

$m$ can't be negative so

$m=8$

General term is

$^{8}C_{r}(x^{16-2r})\bigg(\dfrac{-2}{x}\bigg)^{r}$

$16-3{r}=4\implies r=4$

The term of expansion containing

$x^{4}$ is

$^{8}C_{4}(2)^{4}x^{4}$

Write general term of this:-
$2x{\left( {3 + 2{x^2}} \right)^{20}}$

$2x{\left( {3 + 2{x^2}} \right)^{20}}$

$2x\,\,\,\, \times \,{\,^{20}}{C_r}{\left( 3 \right)^r} \times {\left( {2{x^2}} \right)^{20 - r}}$

Find the middle term of ${ \left( \cfrac { a }{ x } +\cfrac { x }{ a } \right) }^{ 10 }$ .

The expansion of

$(1+x)^n$ has n+1 terms,

$\therefore \ 6^{th}$ term will be the middle term in the expansion

We know that the

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$

$6^{th}$ term in the expansion of

$\left(\cfrac{a}{x}+\cfrac{x}{a}\right)^{10}$ is the middle term which equals

${}^{10}C_{5} \left(\cfrac{a}{x}\right)^5 \left(\cfrac{x}{a}\right)^5=252$

Show that $\displaystyle ^{ n }C_{ 0 } -\ ^{ n }C_{ 1 }+^{ n }C_{ 2 }\ -.....=0$

$(1-x)^{n} = ^{n}C_{0}-^{n}C_{1}x+^{n}C_{2}x^{2}+....^{n}C_{n}x^{4}$

Put

$x = 1$

$0 = ^{n}C_{0}-^{n}C_{1}+^{n}C_{2}+....$

1236854_802825_ans_a185ac07a3e34c23b61637f6d72cc6d0.jpg

Write down and simplify:
The 7th term of ${ \left( \cfrac { 4x }{ 5 } -\cfrac { 5 }{ 2x } \right) }^{ 9 }\quad$

The

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ the

$7^{th}$ term in the expansion of

${\left(\cfrac{4x}{5}-\cfrac{5}{2x}\right)}^{9}$ is

$^{9}C_6 \cfrac{(4x)^3}{{5}^3}\cfrac{(-5)^6}{(2x)^6}$ which equals

$\cfrac{10500}{x^3}$

Find the sum $\displaystyle \sum_{r=1}^{n}\dfrac {r {^{n}C_{r}}}{{^{n}C_{r-1}}}$ .

$\dfrac{^{n}C_{r}}{^{n}C_{r-1}}=\dfrac{\dfrac{n!}{r!(n-r)!}}{\dfrac{n!}{(r-1)!(n-r+1)!}}=\dfrac{n-r+1}{r}$

$\Sigma^{n}_{r=1}r\dfrac{^{n}C_{r}}{^{n}C_{r-1}}=\Sigma^{n}_{i=1}(n-r+1)=(n+1)\Sigma^{n}_{i=1}1-\Sigma^{n}_{i=1}n=n(n+1)-\dfrac{n(n+1)}{2}=\dfrac{n(n+1)}{2}$

Find the $7 ^ { th }$ term of $\left( 3 x ^ { 2 } - \frac { 1 } { 3 } \right) ^ { 10 }$ .

$7_{7}= ^{10}C_{6}(3x^{2})^{4}(\frac{-1}{3})^{6}$

$= \frac{10!}{6!4!}.\frac{81.x^{8}}{729}$

$= \frac{10\times 9\times 8\times 7}{9\times 3\times 2}\times \frac{81x^{8}}{729} = \frac{70x^{8}}{3}$

1195815_1247949_ans_a1e4c603c7664661a45402c8cb005189.jpg

If $4^{th}$ term in the expansion of $\left(ax+\dfrac{1}{x}\right)^n$ is $\dfrac{5}{2}$ , then find the value of $a$ and $n$ .

1516847_1254933_ans_d89c2370312648159e0d88dfaa42e882.jpg

Find the coefficient of $x^5$ in the expansion of the product $(1+2x)^6(1-x)^7$ .

1339958_1131724_ans_28eff8a81ac64b1fbdccfabba37b649e.jpg

$^{ 15 }{ C_{ 3 } }{ +^{ 15 } }{ C_{ 5 } }+.......{ +^{ 15 } }{ C_{ 15 } }$

$\begin{array}{l} According\, to\, question, \\ \, \, { \, ^{ 15 } }{ C_{ 3 } }{ +^{ 15 } }{ C_{ 5 } }+.......{ +^{ 15 } }{ C_{ 15 } } \\ \, we\, know\, that:\, \, { \, ^{ 15 } }{ C_{ 1 } }{ +^{ 15 } }{ C_{ 3 } }{ +^{ 15 } }{ C_{ 5 } }+.......{ +^{ 15 } }{ C_{ 15 } }={ 2^{ 15-1 } }={ 2^{ 14 } } \\ \, so,that\, \\ { \Rightarrow ^{ 15 } }{ C_{ 3 } }{ +^{ 15 } }{ C_{ 5 } }+.......{ +^{ 15 } }{ C_{ 15 } }={ 2^{ 14 } }-15\, \, \quad is\quad Answer. \end{array}$

Show that the middle term in the expansion of $(1 + x)^{n}$ is $6x^2$ if $n=4$

The middle term of expansion when n is even is

$^nC_{n/2}x^{n/2}$

Here

$n=4$

So the term is

$^4C_{4/2}x^{4/2}=6x^2$

The sum of the rational terms in the expansion of $(\sqrt{2}+3^{1/5})^{10}$ is _______.

Let

$T_{r+1}$ be the general term in the expansion of

$(\sqrt 2+3^{1/5})^{10}$

$\therefore \ T_{r+1}=\ ^{10}C_1 (\sqrt 2)^{10-r} (3^{1/5})^r (0\le r \le 10)$

$=\dfrac {10!}{r! (10-r)!}2^{5-r/2} 3^{r/5}$

$T_{r+1}$ will be rational if

$2^{5-r/2}$ and

$3^{r/5}$ are rational numbers. Hence

$5-r/2$ and

$r/5$ are integers. So,

$r=0$ and

$r=10$ . Therefore,

$T_1$ and

$T_{11}$ are rational terms. Now, sum of

$T_1$ and

$T_{11}$ is

$^{10}C_0 2^{5-0}\times 3^0 +\ ^{10}C_{10}2^{5-5}\times 3^2 =32+9=41$ .

The sum of the coefficient of the polynomial $(1+x-3x^{2})^{2163}$ is______

If we put

$x=1$ in the expansion of

$(1+x-3x^2)^{2163}=A_0+A_1x+A_2x^2+...$ we will get the sum of coefficients of the given polynomial, which is equal to

$-1$ .

Find the coefficient of $x^{2}$ in $(x^{2}+\dfrac{1}{x^{3}})^{6}$ .

Let

$T_{r+1}$ be the term containing

$x^{2}$ .

$T_{r+1} = {6_{C}}_{r} (x^{2})^{6-r} (\dfrac{1}{x^{3}})^{r}$

$={6_{C}}_{r}x^{12-5r}$

As the coefficient of x is 2.

$12-5r=2$

$r=2$

Therefore, coefficient of

$x^{2}={6_{C}}_{2}=15$

If the middle term in the expansion of $\left(\dfrac{x}{2}+2\right)^{8}$ is $1120$ ; then the sum of possible real values of $x$ is

Middle term is

$\left(\dfrac {n}{2}+1\right)^{th}$ , i.e.,

$(4+1)^{th}$ , i.e.,

$T_{5}$

$\therefore \ T_5 =^8C_4 \left(\dfrac x2 \right)^4. 2^4 =1120\ \Rightarrow x^4 =\dfrac {8.7.6.5}{1.2.3.4}x^4 =1120$

$\Rightarrow \ x$ =\dfrac {1120}{70}=16$

$\Rightarrow \ (x^2 +4)(x^2 -4)=0$

$\therefore \ x=\pm 2$ only as

$x\ \in R$

Find the middle terms in the expansion of $\left(x^{2}+\dfrac {1}{x}\right)^{7}$ .

General term,

$T_{r+1}={^7C_r}(x^2)^r\left(\dfrac{1}{x}\right)^{7-r}={^7C_r}\cdot x^{2r-7+r}$

$={^7C_r.x^{3r-7}}$

As total number of terms are

$(n+1)=8$

$\Rightarrow$ Middle terms would be

$\left(\dfrac{n}{2}\right.$ and

$\left.\dfrac{n}{2}+1\right)^{th}$

$\Rightarrow 4^{th}$ and

$5^{th}$

$\Rightarrow 4^{th}$ term

$r=3\Rightarrow T_4={^{7}C_3.x^2}$

$5^{th}$ term

$r=4$

$\Rightarrow T_5={^7C_4x^5}$ .

1216072_1396808_ans_0bf36a6f705548b9a3ba88ed595e76ee.jpg

Find the coefficient of $x^3$ in the expansion of $(2+5x)(1-2x)^2$ .

$(2+5x)(1-2x)^2\\(2+5x)(1-4x+4x^2)\\2-8x+8x^2+5x-20x^2+20x^3\\2-3x-12x^2+20x^3$

The coefficient of

$x^3$ is

$20$

The number of integral terms in the expansion of ${ \left( { 5 }^{ 1/2 }+{ 7 }^{ 1/8 } \right) }^{ 1024 }$ is................

${ ({ 5 }^{ \frac { 1 }{ 2 } }+{ 7 }^{ \frac { 1 }{ 8 } }) }^{ 1024 }\\ General\quad Term\quad for\quad the\quad above\quad series\quad is\quad \\ { T }_{ r+1 }=^{ 1024 }C_{ r }{ ({ 5 }^{ \frac { 1 }{ 2 } }) }^{ r }({ { 7 }^{ \frac { 1 }{ 8 } }) }^{ 1024-r }\\ For\quad the\quad term\quad to\quad be\quad intergral\quad r\quad must\quad be\quad the\quad intergal\quad mulitiple\quad of\quad 8\\ So\quad r=0,8.......,1024\\ 1024=0+(n-1)\times 8\\ n=129\quad \\ So\quad the\quad number\quad of\quad intergal\quad terms\quad are\quad 129.$

If the second term in the expansion ${ \left( { a }^{ 1/13 }+{ a }^{ 3/2 } \right) }^{ n }$ is $14{ a }^{ 5/2 }$ , then the value of $\cfrac { { _{ }^{ n }{ C } }_{ 3 } }{ { _{ }^{ n }{ C } }_{ 2 } }$ is .........

Second Term of the expansion is

${ T }_{ 2 }=^{ n }C_{ 1 }\quad \times { (a) }^{ \dfrac { 1 }{ 13 } }\times { (a) }^{ \dfrac { 3(n-1) }{ 2 } }=14{ a }^{ \dfrac { 5 }{ 2 } }$

So We can compare both Side so

$^{ n }C_{ 1 }=14$ So

$n=14$

To find

$\dfrac { ^{ 14 }C_{ 3 } }{ ^{ 14 }C_{ 2 } } =\dfrac { 14-3+1 }{ 3 } =4$ [Note :-Important Formula

$\dfrac { ^{ n }C_{ r } }{ ^{ n }C_{ r-1 } } =\dfrac { (n-r+1) }{ r } ]$

So Answer is

$4.$

If $n$ is odd natural number, then $\sum _{ r=0 }^{ n }{ \cfrac { { \left( -1 \right) }^{ r } }{ { _{ }^{ n }{ C } }_{ r } } }$ equals................

$\displaystyle \sum _{ r=0 }^{ n }{ \cfrac { { \left( -1 \right) }^{ r } }{ { ^{ n }{ C } }_{ r } } } =\frac { 1 }{ { ^{ n }{ C } }_{ 0 } } -\frac { 1 }{ { ^{ n }{ C } }_{ 1 } } +\frac { 1 }{ { ^{ n }{ C } }_{ 2 } } +.......+\frac { 1 }{ { ^{ n }{ C } }_{ n-1 } } -\frac { 1 }{ { ^{ n }{ C } }_{ n } }$ (Since

$n$ is odd natural number)

$\displaystyle \sum _{ r=0 }^{ n }{ \cfrac { { \left( -1 \right) }^{ r } }{ { ^{ n }{ C } }_{ r } } } =\frac { 1 }{ { ^{ n }{ C } }_{ n } } -\frac { 1 }{ { ^{ n }{ C } }_{ n-1 } } +\frac { 1 }{ { ^{ n }{ C } }_{ n-2 } } +.......+\frac { 1 }{ { ^{ n }{ C } }_{ n-1 } } -\frac { 1 }{ { ^{ n }{ C } }_{ n } }$ (

$\because { ^{ n }{ C } }_{ r }={ ^{ n }{ C } }_{ n-r }$ )

$=0$

Coefficient of $x^3$ and $x^4$ in ${ \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 } \right) }^{ 199 }{ \left( x-1 \right) }^{ 201 }$ are .

$S=(1+x+x^{2}+x^{3}+x^{4})^{199}(x-1)^{201}$

$=(\dfrac{x^{5}-1}{x-1})^{199}(x-1)^{201}$

$=(x^{5}-1)^{199}(x-1)^{2}$

$=(1-2x+x^{2})(x^{5}-1)^{199}$

$=(x^{5}-1)^{199}-2x(x^{5}-1)^{199}+x^{2}(x^{5}-1)^{199}$ .
Hence there will be no term of

$x^{3}$ and

$x^{4}$ .

2.

(A)

${ S }_{ n }=4+11+22+37...{ T }_{ n }$

$\displaystyle { T }_{ n }=4+\frac { \left( n-1 \right) }{ 2 } \left( 14+\left( n-2 \right) 4 \right) =1+{ 2n }^{ 2 }+n.$
(B)

$| { 1 }^{ 2 }-{ 2 }^{ 2 }+{ 3 }^{ 2 }-{ 4 }^{ 2 }...2n.$ times

$|$

$=|\left( 1-2 \right) \left( 1+2 \right) +\left( 3-4 \right) \left( 3+4 \right) +\left( 5-6 \right) \left( 5+6 \right) ...2n$ times

$|$

$=|-3-7-11-15...2n$ times

$|$

$=3+7+1...n$ times

$\displaystyle =\left( \frac { n }{ 2 } \left( 6+\left( n-1 \right) 4 \right) \right) =\frac { n }{ 6 } \left( 4n+2 \right) =\left( { 2n }^{ 2 }+n \right)$
(C)

$=3+7+11+15...={ 2n }^{ 2 }+n.$
(D) coffeient of

${ x }^{ n }$ is

$-2(1+2+3+...n$ time

$)$

$\displaystyle =\frac { 2n\left( n+1 \right) }{ 2 } =-{ n }^{ 2 }+n$

If the sum of the coefficients in the expansion of ${(x+y)}^{n}$ is $4096$ , find the greatest coefficient in the expansion.

Sum of all the coefficient of

$(x+y) ^{ n }$ is

$4096$

Sum

$={ 2 }^{ n }=4096$

When

$x=1$ and

$y=1$

$n=12$ .

[Note: Greatest Coefficient of

$(1+x) ^{ n }$ is the Coefficient of Middle

Term

$^{ n }C_{\frac{n}{2} }$ , if

$n$ is even and if

$n$ is odd then

$^{ n }C_{\frac{(n+1)}{2}}$ or

$^{ n }C_{\frac{(n-1)}{2}}]$

So, the greatest coefficient is

$^{ 12 }C_{ 6 }=924$ .

The greatest binomial coefficient in the expansion of $\displaystyle \left ( a+b \right )^{n}$ is $\displaystyle ^{k}C_{m}$ given that the sum of all the coefficients is equal to $4096$ . Find $k-m$ ?

Let

$a=b=1$
Hence sum of coefficients

$=2^{n}=4096$
Hence

$2^{n}=2^{12}$

$\Rightarrow n=12$
The greatest coefficient will be the coefficient of the middle term

$=\:^{12}C_{\tfrac{12}{2}}$

$=\:^{12}C_{6}$ .

Following question contains statements given in two columns which have to be matched The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s Any given statement in Column-I can have correct matching with ONE statement in Column-II
Column-I Column-II

(A)

$\displaystyle \left ( 2n+1 \right )\left ( 2n+3 \right )\left ( 2n+5 \right ).....\left ( 4n-1 \right )$

$\displaystyle \frac{\left ( 2n! \right )\left ( 2n+1 \right )\left ( 2n+2 \right )\left ( 2n+3 \right )\left ( 2n+4 \right )....\left ( 4n-1 \right )\left ( 4n \right )}{\left ( 2n! \right )\left ( 2n+2 \right )\left ( 2n+4 \right )\left ( 2n+6 \right )....\left ( 4n \right )}$

$\displaystyle \frac{\left ( 4n! \right )\left ( n! \right )}{\left ( n! \right )\left ( 2n \right )!2^{n}\left ( n+1 \right )\left ( n+2 \right )...\left ( 2n \right )}$

$\displaystyle \frac{\left ( 4n! \right )\left ( n! \right )}{2^{n}\cdot \left ( 2n \right )!\left ( 2n \right )!}$
(B)

$\displaystyle\sum_{r=1}^{n}r\frac{\hat{n}C_{r}}{\hat{n}C_{r-1}}=\frac{r.n!}{r!\left ( n-r \right )!}\frac{\left ( r-1 \right )!\left ( n-r+1 \right )!}{n!}=\sum_{r==1}^{n}\left ( n-r+1 \right )=n\left ( n+1 \right )-\frac{n\left ( n+1 \right )}{2}=\frac{n\left ( n+1 \right )}{2}$
(C)

$\displaystyle \left ( C_{0}+C_{1} \right )\left ( C_{1}+C_{2} \right )\left ( C_{2}+C_{3}\right ).....\left ( C_{n-1}+C_{n} \right )=mC_{1}C_{2}...C_{n-1}$

$\displaystyle \left ( \frac{C_{0}}{C_{1}}+1 \right )\left ( \frac{C_{1}}{C_{2}}+1 \right )\left ( \frac{C_{2}}{C_{3}}+1 \right ).....\left ( \frac{C_{n-1}}{C_{n}}+1 \right )=m$

$\displaystyle \frac{n+1}{n}\cdot \frac{n+1}{n-1}\cdot \frac{n+1}{n-2}....\frac{n+1}{1}=m$

$\displaystyle \frac{\left ( n+1 \right )^{n}}{n!}=m$
(D)

$\displaystyle \sum_{i=1}^{n}\left \{ \sum_{j=1}^{n}iC_{i}C_{j}+\sum_{j=1}^{n}jC_{i}C_{j} \right \}$

$\displaystyle \sum_{i=1}^{n}iC_{i}\left ( 2^{n}-1 \right )+\sum_{i=1}^{n}C_{i}\cdot 2^{n-1}$

$\displaystyle =n.2^{n-1}\left ( 2^{n} -1\right )+n\cdot 2^{n-1}\left ( 2^{n}-1 \right )=n\cdot 2^{n}\left ( 2^{n}-1 \right )$

If $\displaystyle \left ( 1\times x \right )\left ( 1 +x+x^{2} \right )\left ( 1+x+x^{2}+x^{3} \right ).....\left ( 1+x+x^{2}+x^{3}+......+x^{n} \right )\equiv$
$a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+.......a_{m}x^{m}$ then $\displaystyle \sum_{r=0}^{m}a_{r}$ has the value equal to $\displaystyle \left ( n+k \right )!$ .Find $k$ ?

For the sum of the coefficients, we substitute x=1.
Then we get

$2\times 3\times 4...\times n\times (n+1)$

$=(n+1)!$ .
Hence

$k=1$ .

Find the negative of middle term in the expansion of
$\displaystyle \left ( \frac{2x}{3}-\frac{3}{2x} \right )^{6}$

The middle term will be

$T_{4}$

$=T_{3+1}$

$=-\:^{6}C_{3}.\dfrac{2}{3}^{3}.(\dfrac{3}{2})^{3}.x^{0}$

$=-\:^{6}C_{3}$

$=-20$

If $\displaystyle ^{n+1}C_{2}+2\left ( ^2C_{2}+^3C_{2}+^4C_{2}+.....+^nC_{2} \right )=1^{2}+2^{2}+3^{2}+....+100^{2}$ then find sum of digits of n ?

$\:^{n+1}C_{2}+2[\:^{2}C_{2}+\:^{3}C_{2}+\:^{4}C_{2}+...\:^{n}C_{2}]$

$=\:^{n+1}C_{2}+2[\:^{3}C_{3}+\:^{3}C_{2}+\:^{4}C_{2}+...\:^{n}C_{2}]$

$=\:^{n+1}C_{2}+2[\:^{n+1}C_{3}]$

$=\:^{n+2}C_{3}+\:^{n+1}C_{3}$

$=\dfrac{(100)(2(100)+1)(100+1)}{6}$

$=\dfrac{100\times(201)\times(101)}{6}$
This is true for n=100.

Number of terms in the expansion of $(x^{1/3} + x^{2/5})^{40}$ with integral power of x is equal to

$T_{r + 1} = ^{40} C_r \displaystyle x^{\displaystyle \frac{40 - r}{3}}. x^{\displaystyle \frac{2r}{5}}$

$= ^{40} C_r x^{\displaystyle \frac{200 + r}{15}}$

$r = 10, 25, 40$

$\Rightarrow$ Number of such terms = 3

Find the middle term in the expansion of $(5x\, -\, 7y)^7$ .

Since n is an odd number, the expansion contains two middle terms.

$\left ( \displaystyle \frac {7\, +\, 1}{2} \right )^{th}$ and

$\left ( \displaystyle \frac {7\, +\, 3}{2} \right )^{th}$ terms are the two middle terms in the expansion of (5x - 7y)

$^7$

$T_4\, =\, T_{3 + 1}\, =\, (-1)^3\, ^7C_3\, (5x)^{7 - 3}\, (7y)^3$

$=\, -^7C_3\, (5x)^4\, (7y)^3$

$T_5\, =\, T_{4 + 1}\, =\, (-1)^4\, ^7C_4\, (5x)^{7 - 4}\, (7y)^4$

$=\, -^7C_4\, (5x)^3\, (7y)^4$

Find the 7th term of $\left (\dfrac {4x}{5}-\dfrac {5}{2x}\right )^9$ .

7th term of

$\left (\dfrac {4x}{5}-\dfrac {5}{2x}\right )^9$

$T_{6+1}={}^9C_6\left (\dfrac {4x}{5}\right )^{9-6} \left (-\dfrac {5}{2x}\right )^6$

$=\dfrac {9!}{3!6!}\left (\dfrac {4x}{5}\right )^3\left (\dfrac {5}{2x}\right )^6=\dfrac {10500}{x^3}$ .

Find the number of rational terms in the expansion of $\left ( 9^{1/4}+8^{1/6} \right )^{1000}.$

The general term in the expansion of

$\left ( 9^{1/4}+8^{1/6} \right )^{1000}$ is

$\displaystyle T_{r+1}={}^{1000}\textrm{C}_{r}\left ( 9^{\frac{1}{4}} \right )^{1000-r}\left ( 8^{\frac{1}{6}} \right )^{r}={}^{1000}\textrm{C}_{r}3^{\frac{1000-r}{2}}2^{\frac{r}{2}}$
The above term will be rational if exponent of

$3$ and

$2$ are integers
It means

$\displaystyle \frac{1000-r}{2}$ and

$\displaystyle \frac{r}{2}$ must be integers
The possible set of values of

$r$ is

$\left \{ 0, 2, 4, ..., 1000 \right \}$
Hence, number of rational terms is

$501$ .

$\displaystyle \left ( 3a-\frac{a^{3}}{6} \right )^{9}$

$\displaystyle \left ( 3a-\frac{a^{3}}{6} \right )^{9}$
Here,

$n$ is odd therefore, middle terms are

$\displaystyle \left ( \frac{9+1}{2} \right )$ th &

$\displaystyle \left ( \frac{9+1}{2}+1 \right )$ th
It means

$T_{5}$ &

$T_{6}$ are middle terms

$\displaystyle T_{5}=^{9}\textrm{C}_{4}\left ( 3a \right )^{9-4}\left ( -\frac{a^{3}}{6} \right )^{4}=\frac{189}{8}a^{17}$

$\displaystyle T_{6}=^{9}\textrm{C}_{5}\left ( 3a \right )^{9-5}\left ( -\frac{a^{3}}{6} \right )^{5}=\frac{21}{16}a^{19}.$

Find the middle term in the expansion of $(2x + 3y)$ $^8$ .

Since n is even number,

$\left ( \displaystyle \frac {8}{2}\, +\, 1 \right )^{th}$
term i.e., 5th term is the middle term in (2x + 3y)

$^8$ .

$T_5\, =\, T_{4 + 1}\, =\, ^8C_4\, (2x)^{8 - 4}\, (3y)^4\, =\, ^8C_4\, (2x)^4\, (3x)^4$

Find the middle term(s) in the expansion of:
$\left (1-\dfrac {x^2}{2}\right )^{14}$

$\left (1-\dfrac {x^2}{2}\right )^{14}$
Here,

$n$ is even, therefore middle term is

$\left (\dfrac {14+2}{2}\right )th$ term.
It means

$T_8$ is the middle term

$T_8={}^{14}C_7\left (-\dfrac {x^2}{2}\right )^7=-\dfrac {429}{16}x^{14}$ .

Find the middle term(s) in the expansion of
$\displaystyle \left ( 1-\frac{x^{2}}{2} \right )^{14}$

$\displaystyle \left ( 1-\frac{x^{2}}{2} \right )^{14}$
Here,

$n$ is even, therefore middle term is

$\displaystyle \left ( \frac{14+2}{2} \right )$ th term.
It means

$T_{8}$ is middle term

$\displaystyle T_{8}={}^{14}\textrm{C}_{7}\left ( -\frac{x^{2}}{2} \right )^{7}=-\frac{429}{16}x^{14}$

Find the $13^{th}$ term in the expansion of $\left ( 9x - \dfrac{1}{3\sqrt x} \right )^{18}, x \neq 0$

It is known that

$(r+1)$ term

$(T_{r+1})$ in the binomial expansion of

$(a+b)^n$ is given by

$T_{r+1}=^nC_r a^{n-r}b^r$
Thus

$13^{th}$ term in the expansion of

$\left ( 9x - \dfrac{1}{3\sqrt x} \right )^{18}$ is

$T_{13}= T_{12+1}=\displaystyle ^{18}C_{12}(9x)^{18-12}\left (- \frac{1}{3\sqrt x} \right)^{12}$

$\displaystyle =(-1)^{12} \frac{18!}{12!6!}(9)^6 (x)^6 \left ( \frac{1}{3} \right )^{12}\left ( \frac{1}{\sqrt x} \right )^{12}$

$\displaystyle = \frac{18.17.16.15.14.13.12!}{12!.6.5.4.3.2}.x^6\left ( \frac{1}{x^6} \right ).3^{12}\left ( \frac{1}{3^{12}} \right )$

$=18564$

Write the general term in the expansion of $(x^2 - y)^6$

It is known that the general term $(T)_{r+1}$ {which is the $(r+1)$ in the

binomial expansion of $(a+b)^{n}$ is given by $T_{r+1} = ^nC_r a^{n-r}b^r$

Thus, the general term in the expansion of $(x^2 - y)^6$ is

$T_{r+1} = ^6C_r (x^2)^{6-r} (-y)^r = (-1)^r 6C_r x^{12-2r} y^r$

Find the $4^{th}$ term in the expansion of $(x-2y)^{12}$

It is known that

$(r+1)$ th term

$(T_{r+1})$ in the binomial expansion of

$(a+b)^n$ is given by

$T_{r+1} = ^nC_r b^{n-r}b^r$
Thus the 4th term in the expansion of

$(x+2y)^{12}$ is

$\displaystyle T_4 = T_{3+1}= ^ {12}C_3(x)^{12-3}(-2y)^3 = (-1)^3 \frac{12!}{3!9!}.x^9.(2)^3.y^3$

$\displaystyle \quad = - \frac{12.11.10}{3.2}.(2)^3x^9y^3=-1760x^9y^3$

Find the middle terms in the expansion of $\left (3- \dfrac{x^3}{6} \right )^7$

It is known that in the expansion of

$(a+b)^n$ if n is odd, then there are two middle terms namely

$\left (\dfrac{n+1}{2} \right )^{th}$ term and

$\left (\dfrac{n+1}{2} +1 \right )^{th}$ term.
Therefore, the middle terms in the expansion of

$\left (3- \dfrac{x^3}{6} \right )^7$ are

$\left (\dfrac{7+1}{2}\right)=4^{th}$ term and

$\left (\dfrac{7+1}{2} +1 \right)=5^{th}$ term.

$T_4 = T_{3+1}= ^7C_3(3)^{7-3}\left ( -\dfrac{x^3}{6} \right )^3= (-1)^3 \dfrac{7!}{3!4!}.3^4.\dfrac{x^9}{6^3}$

$= - \dfrac{7.6.5.4!}{3.2.4!}.3^4.\dfrac{1}{2^3.3^3}.x^9=-\dfrac{105}{8}x^9$

$T_5 = T_{4+1} = ^7C_4(3)^{7-4}\left ( -\dfrac{x^3}{6} \right )^4= (-1)^4 \dfrac{7!}{4!3!}.(3)^3.\dfrac{x^{12}}{6^4}$

$= \dfrac{7.6.5.4!}{4!3.2}\dfrac{3^3}{2^4.3^4}.x^{12}=\dfrac{35}{48}x^{12}$
Thus, the middle terms in the expansion of

$\left (3- \dfrac{x^3}{6} \right )^7$ are

$-\dfrac {105}{8}x^9$ and

$\dfrac {35}{48}x^{12}$

Find the middle terms in the expansion of $\left (\dfrac {x}{3}+ 9y\right)^{10}$

It is known that in the expansion

$(a+b)^n$ if

$n$ is even, then the middle terms is

$\left (\dfrac {n}{2}+ 1\right)^{th}$ term.
Therefore, the middle term in the expansion of

$\left (\dfrac {x}{3}+ 9y\right)^{10}$ is

$\left (\dfrac {10}{2}+ 1\right)^{th}$ term
which is,

$\displaystyle T_6 = T_{5+1}= ^{10}C_5 \left (\dfrac{x}{3}\right )^{10-5}(9y)^5= \frac{10!}{5!5!}.\frac{x^5}{3^5}.9^5.y^5$

$\displaystyle \quad =\frac{10.9.8.7.6.5!}{5.4.3.2.5!} \dfrac{1}{3^5}.3^{10}.x^5y^5$

$\quad =252 \times 3^5.x^5.y^5 = 61236x^5y^5$
Thus the middle term in the expansion of

$\left (\dfrac {x}{3}+ 9y\right)^{10}$ is

$61236x^5y^5$

Write the middle terms in the expansion of $\left(\dfrac{3x}{7}-2y\right)^{10}$ .

Fact: If

$n$ is even number then then middle term in the expansion of

$(x+y)^n$ will be

$^nC_{\frac{n}{2}} x^{\frac{n}{2}}y^{\frac{n}{2}}$

So middle term in the expansion of

$\left(\dfrac{3x}{7}-2y\right)^{10}$ is

$^{10}C_5 \left(\dfrac{3x}{7}\right)^5(-2y)^5$

$=\left(\dfrac{6}{7}\right)^5$

$^{10}C_5(-xy)^5$

If $P$ and $Q$ are the sum of odd terms and the sum of even terms respectively, in the expansion of $(x + a)^{n}$ then prove that
(i) $P^{2} - Q^{2} = (x^{2} - a^{2})^{n}$
(ii) $4PQ = (x + a)^{2n} - (x - a)^{2n}$

$(a+x)^{n}=\:^{n}C_{0}a^{n}+\:^{n}C_{1}a^{n-1}x+\:^{n}C_{2}a^{n-2}x^{2}+...\:^{n}C_{n}x^{n}$
And

$(a-x)^{n}=\:^{n}C_{0}a^{n}-\:^{n}C_{1}a^{n-1}x+\:^{n}C_{2}a^{n-2}x^{2}+...(-1)^{n}\:^{n}C_{n}x^{n}$
Adding, we get

$(a+x)^{n}+(a-x)^{n}=2[\:^{n}C_{0}a^{n}+\:^{n}C_{2}a^{n-2}x^{2}+\:^{n}C_{4}a^{n-4}x^{4}+\:^{n}C_{6}a^{n-6}x^{6}+...\:^{n}C_{n}x^{n}]$
Hence

$P=\dfrac{(a+x)^{n}+(a-x)^{n}}{2}$ ...(i)
Similarly

$Q=\dfrac{(a+x)^{n}-(a-x)^{n}}{2}$ ...(ii)
Therefore

$PQ=\dfrac{(a+x)^{n}+(a-x)^{n}}{2}\times\dfrac{(a+x)^{n}-(a-x)^{n}}{2}$

$=\dfrac{(a+x)^{2n}-(a-x)^{2n}}{4}$
Or

$4PQ=(a+x)^{2n}-(a-x)^{2n}$
And

$P^2-Q^2$

$=(P+Q)(P-Q)$

$=\left[\dfrac{(a+x)^{n}+(a-x)^{n}}{2}+\dfrac{(a+x)^{n}-(a-x)^{n}}{2}\right]\left[\dfrac{(a+x)^{n}+(a-x)^{n}}{2}-\dfrac{(a+x)^{n}-(a-x)^{n}}{2}\right]$

$=(a+x)^{n}.(a-x)^{n}$

$=(a^{2}-x^{2})^{n}$
Hence proved.

Prove that $C_0 +C_1 +C_2 +C_3 + ...... + 2^n . C_n = 3^n$ .

Fact:

$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+^nC_3x^3+............+^nC_nx^n$

Substitute

$x=2$ in above identity

$(1+2)^n=^nC_0+^nC_1(2)+^nC_2(2)^2+^nC_3(2)^3+............+^nC_n(2)^n$

$\therefore ^nC_0+2\cdot ^nC_1+4\cdot ^nC_2+8\cdot ^nC_3+........+2^n\cdot ^nC_n=3^n$

Hence proved

Find the fifth term of ${ \left( a+{ 2x }^{ 3 } \right) }^{ 17 }$ .

The

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

Similarly, the fifth term in the expansion of

$(a+2x^3)^{17}$ will be given by

$^{17}C_4a^{13}(2x)^4$

$=\cfrac{17.16.15.147}{1.2.3.4}\times 16a^{13}x^{12}=38080a^{13}x^{12}$

Write down and simplify:
The 10th term of ${ \left( 1-2x \right) }^{ 12 }$

We know that the

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ the tenth term in expansion of

$(1-2x)^{12}$ is

$^{12}C_3(-2x)^9=-112640x^9$

Find the fourteenth term of ${ \left( 3-a \right) }^{ 15 }$ .

The

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ The fourteenth term in the expansion of

$(3-a)^{15}$ will be

$^{15}C_{3} (3)^{2} (-a)^{13}= {}^{15}C_2\times (-9a^{13})$

We get the fourteenth term equal to

$-945a^{13}$

Find the middle term of ${ \left( 1-\cfrac { { x }^{ 2 } }{ 2 } \right) }^{ 14 }$ .

The number of terms in the given expression will be

$14+1 = 15 \therefore$ the middle term will be the

$8^{th}$ term
By binomial expansion, the

$8^{th}$ term will be

$^{14}C_7 \left(-\cfrac{x^2}{2}\right)^7 = \cfrac{-429}{16}x^{14}$

Expand $\cfrac { 7+x }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) }$ in ascending powers of $x$ and find the general term.

Assume

$\cfrac { 7+x }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } =\cfrac { A }{ 1+x } +\cfrac { Bx+C }{ 1+{ x }^{ 2 } }$

$\therefore$

$7+x=A(1+{x}^{2})+(Bx+C)(1+x)$
Let

$1+x=0$ , then

$A=3$ ;
Equating the absolute terms,

$7=A+C$ , whence

$C=4$
Equating the coefficients of

${x}^{2}$ ,

$0=A+B$ , whence

$B=-3$

$\therefore \quad \cfrac { 7+x }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } =\cfrac { 3 }{ 1+x } +\cfrac { 4-3x }{ 1+{ x }^{ 2 } }$

$={ 3\left( 1+x \right) }^{ -1 }+(4-3x){ \left( 1+{ x }^{ 2 } \right) }^{ -1 }$

$=3\left\{ 1-x+{ x }^{ 2 }-......+{ \left( -1 \right) }^{ p }{ x }^{ p }+.... \right\} +(4-3x)\left\{ 1-{ x }^{ 2 }+{ x }^{ 4 }-......+{ \left( -1 \right) }^{ p }{ x }^{ 2p }+.... \right\}$

To find the coefficient of

${x}^{r}$ ;

(1) If

$r$ is even, the coefficient of

${x}^{r}$ in the second series is

$4{(-1)}^{\frac{r}{2}}$ ; therefore in the expansion the coefficient of

${x}^{r}$ is

$3+4{(-1)}^{\frac{r}{2}}$

So, the general term is

$(3+4{(-1)}^{\frac{r}{2}})x^{r}$
(2) If

$r$ is odd, the coefficient of

${x}^{r}$ in the second series is

$-3{(-1)}^{\cfrac{r-1}{2}}$ ; therefore in the expansion the coefficient of

${x}^{r}$ is

$3{(-1)}^{\frac{r+1}{2}}-3$

Thus, the general term is

$(3{(-1)}^{\frac{r+1}{2}}-3)x^{r}$ .

The 2nd, 3rd and 4th terms in the expansion of ${(x+y)}^{n}$ are $240,720, 1080$ respectively; find $x,y,n$ .

General term

$T_{r+1}$ of

$(x+y)^{n}$ is given by

$T_{r+1}={}^{n}C_{r}\ x^{n-r}\ y^{r}$

$\Longrightarrow { T }_{ 2 }={}^{ n }{ C }_{ 1 }\cdot { x }^{ n-1 }\cdot y=240$ .....

$(A)$

${ { T }_{ 3 } }={}^{ n }{ C }_{ 2 }\cdot { x }^{ n-2 }\cdot { y }^{ 2 }=720$

${ { T }_{ 4 } }={}^{ n }{ C }_{ 3 }\cdot { x }^{ n-3 }\cdot { y }^{ 3 }=1080$

$\cfrac { { { T }_{ 3 } } }{ { { T }_{ 2 } } } =\cfrac { \left( n-1 \right) }{ 2 } \times \cfrac { y }{ x } =3$ .......

$\left( i \right)$

$\cfrac { { { T }_{ 4 } } }{ { { T }_{ 2 } } } =\cfrac { \left( n-1 \right) \left( n-2 \right) }{ 3\times 2 } \times \cfrac { { x }^{ 2 } }{ { y }^{ 2 } } =\cfrac { 9 }{ 2 }$ ........

$(ii)$

$\cfrac { { { T }_{ 4 } } }{ { { T }_{ 3 } } } =\cfrac { \left( n-2 \right) }{ 3 } \times \cfrac { y }{ x } =\cfrac { 3 }{ 2 }$ .......

$\left( iii \right)$

Dividing equation

$\left( i \right)$ by equation

$\left(iii\right) :-$

$\cfrac { n-1 }{ 2 } \times \cfrac { 3 }{ n-2 } =2$

$\Rightarrow 3n-3=4n-8$

$\Rightarrow 5=n$

$\Rightarrow \cfrac { y }{ x } =\cfrac { 3 }{ 2 }$ ..... From

$(iii)$

Putting

$y=\cfrac { 3 }{ 2 } x$

${ 5x }^{ 5 }\times \cfrac { 3 }{ 2 } =240$ ........ From

$\left(A\right)$

${ x }^{ 5 }=32$

$\boxed { x=2,y=3,n=5 }$

Write down and simplify:
The 4th term of ${ { \left( \cfrac { a }{ 3 } +9b \right) }^{ 10 } }$ .

The

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ the

$4^{th}$ term in the expansion of

${\left(\cfrac{a}{3}+9b\right)}^{10}$ is

$^{10}C_3 \cfrac{a^7}{3^7}9^3b^3$ which equals

$40a^7b^3$

Show that the middle term in the expansion of ${ \left( 1+x \right) }^{ 2n }$ is
$\cfrac { 1.3.5....(2n-1) }{ |\underline { n } } { 2 }^{ n }{ x }^{ n }$ .

Number of terms

$=2n$ which is even

So, middle term

$=(n+1)^{th}$ term

Hence, we need

$T_{n+1}$

General term of

$(a+b)^n$ is

$T_{r+1}={}_{r}^{n}{C}a^{n-r}b^r$

For

$T_{n+1},$ putting

$n=2r,r=n,a=1,b=x$

$T_{n+1}={}_{n}^{2n}{C}(1)^{2n-n}{x}^n$

$=\cfrac{(2n)!x^n}{n!n!}$

$=\cfrac{(2n)(2n-1)(2n-2)...............\times 2\times 1x^n}{n!n!}$

$=\cfrac{((2n-1)(2n-3).....3\times 1)(2n(2n-2).......\times 2 )x^n}{n!n!}$

Simplifying the expression, the middle term

$=\cfrac{1.3.5........(2n-1)2^nx^n}{n!}$

Find the $(p+2)$ th term from the end in ${ \left( x-\cfrac { 1 }{ x } \right) }^{ 2n+1 }$ .

We know that the

$r^{th}$ term from beginning in expansion of

$(x+a)^n$ is given by

$^nC_{r-1}x^{n+1-r}a^{r-1}$

and

$r^{th}$ term from beginning is

$(n-r+2)^{th}$ term from end

$\therefore$ coefficient of

$(n-r)^{th}$ term can be found by putting

$n-r+2$ instead of

$r$ in above equation, from which we get

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\because \ ^nC_{r-1}=^nC_{n-r+1}$

$\therefore$ The

$r^{th}$ term from end in expansion of

$(x+a)^n$ is

$^nC_{r-1}a^{n+1-r}x^{r-1}$

In the given question the term and power are

$p+2 \ \And \ 2n+1$ respectively,

$\therefore$ Replace

$r\Rightarrow p+2$ and

$n\Rightarrow 2n+1$

On putting it in the equation we get

$(p+2)^{th}$ term of

$\left(x-\cfrac{1}{x}\right)^{2n+1}$ equal to

$^{2n+1}C_{p+1}{\left (-\cfrac{1}{x}\right)}^{2n-p}(x)^{p+1}$

$\because (-1)^{2n+p}=(-1)^p$

$\Rightarrow$ the required coefficient is

${(-1)^{p}} \ ^{2n+1}C_{p+1} x^{2p-2n+1}$

Find the rth term from the end in ${ \left( x+a \right) }^{ n }$ .

We know that the

$r^{th}$ term from beginning in expansion of

$(x+a)^n$ is given by

$^nC_{r-1}x^{n+1-r}a^{r-1}$

and

$r^{th}$ term from beginning is

$(n-r+2)^{th}$ term from end

$\therefore$ coefficient of

$(n-r)^{th}$ term can be found by putting

$n-r+2$ instead of

$r$ in above equation, from which we get

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\because\ {}^nC_{r-1}=^nC_{n-r+1}$

$\therefore$ the

$r^{th}$ term from end in expansion of

$(x+a)^n$ is

$^nC_{r-1}a^{n+1-r}x^{r-1}$

Write down and simplify:
The 25th term of ${ \left( 5x+8y \right) }^{ 30 }$

We know that the the

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ the

$25^{th}$ term in the expansion of

$(5x+8y)^{30}$ is

$^{30}C_{24} (5x)^{6}(8y)^{24}$

Write down and simplify:
The 12th term of ${ \left( 2x-1 \right) }^{ 13 }$ .

We know that the

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ the

$12^{th}$ term in the expansion of

$(2x-1)^{13}$ is

$^{13}C_{11} (2x)^2(-1)^{11}=-312x^2$

Write down and simplify:
The 5th term of ${ \left( 2a-\cfrac { b }{ 3 } \right) }^{ 8 }$ .

The

$r^{th}$ term in expansion of

$(a+x)^n$ is given by

$^nC_{r-1}a^{n+1-r}x^{r-1}$

$\therefore$ the

$5^{th}$ term in the expansion of

${\left(2a-\cfrac{b}{3}\right)}^{8}$ is

$^{8}C_4 (2a)^4 \cfrac{(-b)^4}{3^4}$ which equals

$\cfrac{1120}{81}a^4b^4$

If $(1+x)^n=c_0+c_1x+c_2x^2+....+c_nx^n$ , find the value of $1^2c_1+2^2c_2+3^2c_3+.....+n^2c_n$ .

we can easily show that

$1^2+2^2x+3^2x^2+4^2x^3+...+n^2x^{n-1}+.....=\displaystyle \frac{1+x}{(1-x)^3}$ .
Also

$c_n+c_{n-1}x+....c_2x^{n-2}+c_1x^{n-1}+c_0x^n=(1+x)^n$ .
Multiply together these two results

$;$ then the given series is equal to the coefficient of

$x^{n-1}$ in

$\displaystyle\frac{(1+x)^{n+1}}{(1-x)^3},$ that is, in

$\displaystyle\frac{(2-\overline{1-x})^{n+1}}{(1-x)^3}$ .
The only terms containing

$x^{n-1}$ in this expansion arise from

$2^{n+1}(1-x)^{-3}-(n+1)2^n(1-x)^{-2}+\displaystyle \frac{(n+1)n}{\left\lfloor 2\right.}2^{n-1}(1-x)^{-1};$

$\therefore$ the given series

$=\displaystyle\frac{n(n+1)}{\left\lfloor 2\right.}2^{n+1}-n(n+1)2^n+\frac{n(n+1)}{\left\lfloor 2\right.}2^{n-1}$

$=n(n+1)2^{n-2}$

$2.C_0+5.C_1 +8.C_2 + ...+ (3n+2)C_n = (3n+4)2^{n-1}$

$2C_0+5C_1+8C_2+.............(3n+4)C_n$

$\sum_{r=0}^{n}{(3r+2)C_r}$

$\sum_{r=0}^{n}{3r.C_r+2C_r}$ as for r=0

$3rC_r$ would be 0

$\sum_{r=1}^{n}{3r.\dfrac{n}{r}.^{n-1}{C}_{r-1}+\sum_{r=0}^{n}{}2.C_r}$

$\sum_{r=1}^{n}{3n.^{n-1}{C}_{r-1}}$ +

$\sum_{r=0}^{n}{2.C_r}$ As =

$2^n$ =

$^{n}{C}_{0}+^{n}{C}_{1}+^{n}{C}_{2}...........+^{n}{C}_{n}$

$3n.2^{n-1}+2.2^n$

$3n.2^{n-1}+4.2^{n-1}$

$(3n+4).2^{n-1}$

$\sum_{k=1}^{3n} 6 ^nC_{2\, k-1}(-3)^k$ is equal to:

1968203_693583_ans_5a2c0fa872e54234a3db86836c2ef05f.jpg

Prove that $^{n}C_{r} + ^{n - 1}C_{r} + ^{n - 2}C_{r} + ..... + ^{r}C_{r} = ^{n + 1}C_{r + 1}$ .

To prove:

$^nC_r+{^{n-1}C_r}+{^{n-2}C_r}+......+{^rC_r}={^{n+1}C_{r+1}}$

$^nC_r+{^{n-1}C_r}+{^{n-2}C_r}+.....+{^{r+1}C_r}+{^rC_r}={^{n+1}C_{r+1}}$

$[\because {^{n}C_n}={^{n+1}C_{n+1}}]$

$\Rightarrow {^{n}C_r}+{^{n-1}C_r}+{^{n-2}C_r}+......+{^{r+2}C_r}+({^{r+1}C_r}+{^{r+1}C_{r+1}})$

$\Rightarrow {^{r+2}C_{r+1}}$

[We know that

$^{n+1}C_{r+1}={^nC_{r+1}+{^nC_r}}$ ]

$\therefore ^nC_r+{^{n-1}C_r}+{^{n-2}C_r}+......+{^{r+2}C_r}+{^{r+2}C_{r+1}}={^{n+1}C_{r+1}}$

$\Rightarrow ^nC_r+{^n-1}C_r+{^{n-2}C_r}+......+{^{r+3}C_{r+1}}={^{n+1}C_{r+1}}$

By continuing this, we get

$\Rightarrow ^nC_r+({^{n-1}C_r}+{^{n-1}C_{r+1}})={^{n+1}C_{r+1}}$

$\Rightarrow ^nC_r+{^nC_{r+1}}={^{n+1}C_{r+1}}$

$\Rightarrow {^{n+1}C_{r+1}}={^{n+1}C_{r+1}}$

L.H.S

$=$ R.H.S

[Hence proved].

1178549_827283_ans_049f7880d91449c8a4e481e17a2d62e3.jpg

Write down and simplify $6^{th}$ term in the expansion of $\displaystyle \left( \frac{2x}{3} + \frac{3y}{2} \right)^9$ .

${ T }_{ 6 }$ =

$^{9}{C}_{ 5 }{ \left( \dfrac { 2x }{ { 3 } } \right) }^{4}.{ \left( \dfrac { 3y }{ { 2 } } \right) }^{ 5 }$

$\dfrac { 9\times 8\times 7\times 6\times 5 }{ 5\times 4\times 3\times 2 } \times \dfrac { { 2 }^{ 4 } }{ { 3 }^{ 4 } } .\dfrac { { 3 }^{ 5 } }{ { 2 }^{ 5 } } .{ x }^{ 4 }{ y }^{ 5 }$

${ 3 }^{ 3 }\times7{ x }^{ 4 }{ y }^{ 5 }$

$189{ x }^{ 4 }{ y }^{ 5 }$ .

The last term in $(2^{1/3} \, + \, 2^{-1/2})^n \, is \, \left(\dfrac{1}{3(9)^{1/3}}\right)^{log_33}$ then show that the fifth term is 210

The last term =

$T_{n \, + \, 1}$

$= \, ^nC_n \, (2^{1/3})^0 \, (2^{-1/2})^n \, = \, \left[\dfrac{1}{3 (9)^{1/3}} \right]^{log_3 \, 8}$
or

$2^{-n / 2} \, = \, (3^{-5 / 3})^{log_3 \, 8} \, = \, 3^{log_3 \, 8^{-5 / 3}}$

$\because \, \, \, \, a^{log_a \, n} \, = \, n$
or

$2^{-n / 2} \, = \, 8^{-5 / 3} \, = \, (2^3)^{- 5 / 3} \, = \, 2^{-5} \, \, \, \, \therefore \, \, \, n \, = \ 10$

$\therefore \, \, \, \, T_5 \, = \, ^{10}C_4 \, (2^{1 / 3})^6 \, (2^{-1 / 2})^4$
= 210

$(2^2) \, (2^{-2})$ = 210

If ' $n$ ' is a positive integer and ' $x$ ' is any non-zero number, then prove that
$\displaystyle C_0 + C_1 . \frac{x}{2} + C_2 . \frac{x^2}{3} + ..... + C_n . \frac{x^n}{n+1} = \frac{(1+x)^{n+1} - 1}{(n+1)x}$

$(1+x)^n =C_0 +C_1x+C_2x^2+.....C_nx^n$

$\dfrac {(1+x)^n}{(n+1)} =C_0+\dfrac {C_1x^2}{2}+\dfrac {C_2x^3}{3}+....C_n\dfrac {x^{n+1}}{n+1}+C$

Put

$x=0\ LHS=\dfrac {(1)}{(n+1)}$

$RHS=C$

$\therefore \ C=\dfrac {1}{n+1}$

$\dfrac {(1+x)^{n+1}}{n+1}-C=C_0x+\dfrac {C_1x^2}{2}+ \dfrac {C_2 x^3}{3}+.....\dfrac {C_n x^{n+1}}{x+1}$

$\dfrac {[(1+x)^{n+1}-1]}{(n+1)}=x\left(C_0+\dfrac {C_1 x}{2}+\dfrac {C_2 x^2}{3}+....\dfrac {C_ n x^n}{(n+1)}\right)$

$\therefore \boxed {C_0+C_1\dfrac {x}{2}+C_2\dfrac {x^2}{3}+....\dfrac {C_n x^n}{(n+1)}=\dfrac {((1+x)^{n+1}-1)}{(n+1)x}}$

Find the third term of the expansion of $\displaystyle\, \left ( z^2 + \frac{1}{z} \sqrt[3]{z} \right )^n$ , if the sum of all the binomial coefficients is equal to 2048.

for sum of coefficients put

$z=1$

$(2)^{n}=2048\implies 2^{n}=2^{11}\implies n=11$

Third term is

$^{11}C_{2}(z^{2})^{9}\bigg(\dfrac{1}{z}\sqrt[3]{z}\bigg)^{2}$

$=55 z^{{16}-\cfrac{2}{3}}$

$=55 z^{\dfrac{46}{3}}$

Find the middle term of the expansion of $\displaystyle\, \left ( \sqrt{x} - \frac{1}{x} \right )^6$

middle term will be

$^{6}C_{3}(\sqrt{x})^{3}\bigg(-\dfrac{1}{x}\bigg)^{3}$

$=-20 x^{-3/2}$

Find the second term of the binomial expansion of $\displaystyle\, \left ( \sqrt[13]{a} + \frac{a}{\sqrt{a^{-1}}} \right )^m$ , if $\displaystyle\, C_{3}^{m} : C_{2}^{m} = 4 : 1$

$^{m}C_{3}:^{m}C_{2}=4:1$

$\implies \dfrac{m!}{(m-3)!3!}\times \dfrac{2!(m-2)!}{m!}=4$

$\implies \dfrac{m-2}{3}=4\implies m=14$

Second terms is

$^{14}C_{1}(\sqrt[13]{a})^{13}(a^{3/2})=14{a^{5/2}}$

if $a_0 \, , \, a_1 \, , \, a_2$ ....be the coefficients in the expansion of (1 + x + x $^2)^n$ in ascending powers of x , then prove that
$a_{0}^{2} \, - \, a_{1}^{2} \, + \, a_{2}^{2} \, - \, a_{3}^{2} \, + \, ......+ \, (- 1)^{n \, - \, 1} \, a_{n \, - \, 1}^{2}$
$= \, \dfrac{1}{2}a_n(1 \, -(- 1)^n \, a_n)$

Now

$a_{2n} \, = \, a_0 \, , \, a_{2n \, - \, 1} \, = \, a_1$

$\therefore \, \, 2[ a^2_0 \, + \, a^2_1 \, + \, .... \, + \, (-1)^{n \, - \, 1} a^2_{n - 1} ] \, = \, (-1)^n \, a^2_n \, = \, a_n$

$a^2_0 \, + \, a^2_1 \, + \, .... \, + \, (-1)^{n \, - \, 1} {a_{n - 1}}^2$

$= \, \dfrac {1}{2} \, a_n \, [ 1 \, - \, (-1)^n \, a_n ]$

This prove the second part

If the coefficient of $a^{r-1}, a^r$ and $a^{r + 1}$ in the expansion of $(1 + a)^n$ are in arithmetic progression, prove that $n^2 - n(4r + 1) + 4r^2 - 2 = 0$ .

coefficient of

$ar^{-1}$ in

$(1+a)^n$

$^nC_{r-1}(1)^{n-r+1}(a)^{r-1}$

$a^r=^nC_r(1)^{n-r}(a)^r$

Factors

$a^{r+1}+^nC_{r+1}(1)^{n-r-1}a^{r+1}$

Coefficient of

$a^{r-1}+a^{r+1}=2a^r$

$=^nC_{r-1}(1)^{n-r+1}(a)^{r-1}+^nC_{r+1}(1)^{n-r-1}a^{r+1}$

$=2 \ ^nC_r(a)^r$

$=^nC_{r-1}a^{r-1}+^nC_{r+1}a^{r+1}=2^nC_ra^r$

$=^nC_{r-1}\left(\dfrac{1}{a}\right)+^nC_{r+1}a=^{2n}C_r$

$=\dfrac{n!}{(n-r+1)!\times (r-1)!}\left(\dfrac{1}{a}\right)+\dfrac{n!}{(n-r-1)!\times (r+1)!}=\dfrac{2n!}{(n-r)!r!}$

$=\dfrac{1}{a}\left(\dfrac{1}{n-r+1}!\right)+\dfrac{a}{(n-r-1)!r(r+1)}=\dfrac{2}{(n-r)!\times r}$

$=\dfrac{1}{a}\left(\dfrac{1}{(n-r+1)(n-r)}\right)+\dfrac{a}{r(r+1)}=\dfrac{2}{(n-r)r}$

$=\dfrac{1}{(n-r+1)(n-r)}+\dfrac{a^2}{r(r+1)}=\dfrac{2a}{(n-r)r}$

$r(r+1)+a^2(n-r+1)(n-r)=2a(n-r+1)(r+1)$

Simplify

$=n^2-n(4r+1)+4r^2-2=0$

In the expansion of $(7^{1/3} \, + \, 11^{1/9})^{6561}$ prove that three will be only $730$ term which are free from radicals

$T_{r+1} \, = \, ^{6561}C_r \, (7^{1/3})^{6561-r}\,(11^{1/9})^{r}$

$= \, ^{6561}C_r \, 7^{2187} \,\cdot \, 7^{-r/3}\,\cdot \,11^{r/9},\, 0\leq r\leq 6561$
Now

$T_{r+1}$ will be integral if r/3 and r/9 both are integers in the span

$0\leq r\leq 6561$ . It will be possible if r is multiple of 9. In other words, we have to find the numbers of those numbers lying between 0 to 6561 which are multiple of 9 i.e. those which form an A.P. of common difference is 9.
0,9,18,27, ....,6561
Note that 6561 is divisible by 9. If it were not divisible by 9 we would have chosen just the preceding number which is divisible by 9.

$T_n \, = \, 6561 \, = \, a \, + \, (n-1)d \, = \, 0 \, + \, (n-1)9$

$\therefore \, n \, = \, 1 \, + \, \dfrac{6561}{9} \, = \, 1 \, + \, 729 \, = \, 730$
Hence there will be 730 terms will be free from radicals .

If $a_0 \, , \, a_1 \, , \, a_2$ ....be the coefficients in the expansion of $(1 + x + x$ $^2)^n$ in ascending powers $x$ , then prove that :
$a_r \, = \, a_{2n \, - \, r}$

$(1 \, + \, x \, + \, x^2)^n \, = \, a_0 \, + \, a_1x \, + \, a_2x^2 \, + \, \cdots \, + \, a_{2n}x^{2n} \, = \, \sum_{r \, = \, 0}^{2n} \, a_rx^r$
Replace

$x$ by

$\dfrac 1x$ .

$\left(1 \, + \, \dfrac{1}{x} \, + \, \dfrac{1}{x^2} \right)^2 \, = \, a_0 \, + \, a_1 \, \left(\dfrac{1}{x} \right) \, + \, a_2\left(\dfrac{1}{x^2} \right) \, + \, \cdots$
or

$\, \dfrac{(1 \, + \, x \, + \, x^2)^n}{x^{2n}} \, = \, a_0 \, + \, a_1 \, \left(\dfrac{1}{x} \right) \, + \, \cdots \, + \, a_r\left(\dfrac{1}{x^r} \right) \, + \, \cdots$
or

$(1 \, + \, x \, + \, x^2)^n =\, a_0x^{2n} \, + \, a_1x^{2n \, - \, 1} \, + \, \cdots \, + \, a_rx^{2n \, - \, r} \, + \, \cdots$

$\sum_{r \, = \, 0}^{2n} \, a_rx^r \, = \, \sum_{r \, = \, 0}^{2n} \, a_r^{2n \, + \, r}$ , by (1)(2)
Equating the coefficients of

$x^{2n \, - \, r}$ in both sides, we get

$a_{2n \, - \, r} \, = \, a_r \, or \, a_r \, = \, a_{2n \, - \, r}$ , where 0 < r < 2n

Find the middle term(s) of $(\frac{x^{3/2}y}{2} + \frac{2}{xy^{3/2}})^{13}$

Middle terms are

$^{13}C_{\dfrac{13-1}{2}}\bigg(\dfrac{x^{3/2}y}{2}\bigg)^{6}\bigg(\dfrac{2}{x y^{3/2}}\bigg)^{7},^{13}C_{\dfrac{13+1}{2}}\bigg(\dfrac{x^{3/2}y}{2}\bigg)^{7}\bigg(\dfrac{2}{x y^{3/2}}\bigg)^{6}\implies 2 \ ^{13}C_{6}x^{2}y^{-3/2},\dfrac{1}{2}^{13}C_{7}x^{3/2}y^{-2}$

${ C }_{ 0 }+\dfrac { 3 }{ 2 } .{ C }_{ 1 }+\dfrac { 9 }{ 3 } .{ C }_{ 2 }+\dfrac { 27 }{ 4 } .{ C }_{ 3 }+...........+\dfrac { { 3 }^{ n } }{ n+1 } .{ C }_{ n }=\dfrac { { 4 }^{ n+1 }-1 }{ 3\left( n+1 \right) }$ .Is it true ?If true enter 1 else 0.

Solve: we have Given.

$c_{0}+\frac{3}{2} \cdot c_{1}+\frac{9}{3} c_{2}+\frac{27}{4} c_{3}+\cdots+\frac{3^{n}}{n+1} c_{n}=\frac{4^{n+1}-1}{3(n+1)}$

By binomial expansion

$(1+x)^{n}={}^n{c_{0}}+{}^n{c_{1} x}+{ }^{n} c_{2} x^{2}+\ldots . .^{n} c_{n} x^{n}$

on integrating both sides with respect to

$x$ we get

$\int_{0}^{x}(1+x)^{n} d x=\int\left({}^n{c_0}+{}^n{c_1 x}+{}^nc_{2} x^{2}+\ldots {}^n{c_n} x^{n}\right) d x$

$\Rightarrow \frac{(1+x)^{n+1}-1}{n+1}={}^n{c_0} x+{}^n{c_{1}} \frac{x^{2}}{2}+{}^n{c_{2}} \frac{x^{3}}{3}+\ldots{}^n c_{n} \frac{x^{n+1}}{n+1}$

on putting

$x=3$ we get

$\Rightarrow \frac{4^{n+1}-1}{n+1}=3^{n} c_{0}+\frac{9}{2} {}^n{c_{1}}+\frac{27 }{3} {}^nc_{2}+\ldots . . \frac{3^{n+1}}{n+1} {}^n{c_n}$

$\Rightarrow \frac{4^{n+1}-1}{n+1}=3\left[{ }^{n} c_{0}+\frac{3}{2} {}^n{c_1}+\frac{9}{3} {}^n{c_{2}+\ldots} \ldots \frac{3^{n+1}}{n+1} {}^n{c_ n}\right]$

$\Rightarrow{ }^{n} c_{0}+\frac{3}{2}{ }^{n} c_{1}+\frac{9}{3} {}^n{c_2}+\ldots \frac{3^{n}}{n+1}=\frac{4^{n+1}-1}{3(n+1)}$

Hence It is correct

so, correct answer is 1 .

Find the coefficient of $x^{25}$ in expansion of expression $\displaystyle \sum_{r = 0}^{50} {^{50}C}_r (2x - 3)^r (2 - x)^{50 -r}$ .

Now

$\displaystyle \sum_{r = 0}^{50} {^{50}C}_r (2x - 3)^r (2 - x)^{50 -r}=\{(2x-3)+(2-x)\}^{50}=(x-1)^{50}$ .

From this expression it is clear that the co-efficient of

$x^{25}$ is

$^{50}C_{25}(-1)^{25}=-^{50}C_{25}$ .

Find the middle term:
(i) $\left[3x - \dfrac{2}{x}\right]^{15}$

If n is odd

$\left(\dfrac{n+1}{2}\right)^{th}$ &

$\left(\dfrac{n+1}{2}+1\right)^{th}$ term are middle terms.

$\Rightarrow (3x\dfrac{-2}{x})^{15}$

$\therefore n=odd=15$

Middle terms

$=\left(\dfrac{15+1}{2}\right)^{th}=8^{th}$ term &

$\left(\dfrac{15+1}{2}+1\right)^{th}=9^{th}$ term

Term

$8^{th}$ Here

$a=3x, b=\dfrac{-2}{x}, n=15, r=7$

Tern

$8^{th}$

$={^{15}C_7}(3x)^{15-7}\left(\dfrac{-2}{x}\right)^{7}$

$\Rightarrow {^{15}C_7}3^8x^8\times \dfrac{(-2)^7}{x^7}$

$\Rightarrow -{^{15}C_7}3^8\times 2^7x$ .

Term

$9^{th}$ Here

$a=3x, b=\dfrac{-2}{x}, n=15, r=8$

Term

$9^{th}={^{15}C_8}(3x)^{15-8}\left(\dfrac{-2}{x}\right)^8$

$\Rightarrow {^{15}C_8}3^7\times x^7\times \dfrac{(-2)^8}{x^8}$

$\Rightarrow {^{15}C_8}3^7\times 2^8\times x$ .

1045622_1044233_ans_6f8ddfa6741b47caa347a19c9cf03e33.jpg

Find the coefficient of $a^4$ in the product $(1 + 2a)^4 (2 - a)^5$ using binomial theorem.

$\Rightarrow$ (1+2a)^4$$

$\Rightarrow ^{4}C_1+{^{4}C_1}(2a)+{^{4}C_2}(2a)^2+{^{4}C_3}(2a)^3+{^{4}C_4}(2a)^4$

$\Rightarrow 1+8a+24a^2+32a^3+16a^4$

$(2-a)^5$

$\Rightarrow {^{5}C_0}2^5-{^{5}C_1}2^4(a)+{^{5}C_2}2^3(a)^2-{^{5}C_3}2^2a^3+{^{5}C_4}2a^4-{^{5}C_5}a^5$

$\Rightarrow 32-80a+80a^3+10a^4-a^5$

$\therefore$ Coefficient of

$a^4$ is

$\Rightarrow 1\times 10a^4+(8a)(80a^3)+32a^3(-80a)+16a^4(32)$

$\Rightarrow 10a^4+640a^4-2560a^4+512a^4$

$\Rightarrow -1.398x^4$ .

1043068_1040863_ans_146f97e9c0a64606acb8cb58bcf6e0f0.jpg

Evaluate ${(\sqrt{3}+\sqrt{2})}^{6} -{(\sqrt{3}-\sqrt{2})}^{6}.$

${\left( {\sqrt {\rm{3}} + \sqrt {\rm{2}} } \right)^{\rm{6}}} - {\left( {\sqrt {\rm{3}} - \sqrt {\rm{2}} } \right)^{\rm{6}}}$

Using binomial expansion

$= \left[ {^6{C_0}{{\left( {\sqrt 3 } \right)}^6}{{\left( {\sqrt 2 } \right)}^0}{ + ^6}{C_1}{{\left( {\sqrt 3 } \right)}^5}{{\left( {\sqrt 2 } \right)}^1}{ + ^6}{C_2}{{\left( {\sqrt 3 } \right)}^4}{{\left( {\sqrt 2 } \right)}^2}{ + ^6}{C_3}{{\left( {\sqrt 3 } \right)}^3}{{\left( {\sqrt 2 } \right)}^3}{ + ^6}{C_4}{{\left( {\sqrt 3 } \right)}^2}{{\left( {\sqrt 2 } \right)}^4}{ + ^6}{C_5}{{\left( {\sqrt 3 } \right)}^1}{{\left( {\sqrt 2 } \right)}^5}{ + ^6}{C_6}{{\left( {\sqrt 3 } \right)}^0}{{\left( {\sqrt 2 } \right)}^6}} \right] - \left[ {^6{C_0}{{\left( {\sqrt 3 } \right)}^6}{{\left( { - \sqrt 2 } \right)}^0}{ + ^6}{C_1}{{\left( {\sqrt 3 } \right)}^5}{{\left( { - \sqrt 2 } \right)}^1}{ + ^6}{C_2}{{\left( {\sqrt 3 } \right)}^4}{{\left( { - \sqrt 2 } \right)}^2}{ + ^6}{C_3}{{\left( { - \sqrt 3 } \right)}^3}{{\left( {\sqrt 2 } \right)}^3}{ + ^6}{C_4}{{\left( { - \sqrt 3 } \right)}^2}{{\left( {\sqrt 2 } \right)}^4}{ + ^6}{C_5}{{\left( {\sqrt 3 } \right)}^1}{{\left( { - \sqrt 2 } \right)}^5}{ + ^6}{C_6}{{\left( {\sqrt 3 } \right)}^0}{{\left( { - \sqrt 2 } \right)}^6}} \right]$

$= 2\left[ {^6{C_1}{{\left( {\sqrt 3 } \right)}^5}{{\left( {\sqrt 2 } \right)}^1}{ + ^6}{C_3}{{\left( {\sqrt 3 } \right)}^3}{{\left( {\sqrt 2 } \right)}^3}{ + ^6}{C_5}{{\left( {\sqrt 3 } \right)}^1}{{\left( {\sqrt 2 } \right)}^5}} \right]$

$= 2\left[ {6 \times 9\sqrt 6 + 20 \times 6\sqrt 6 + 6 \times 4\sqrt 6 } \right]$

$=396 \sqrt 6$

In the expansion of ${\left( {{x^3} - \dfrac{1}{{{x^2}}}} \right)^2}$ , where n is a positive integer, the sum of the coefficients of ${x^6}$ is $1.$

$\bigg(x^{3}-\dfrac{1}{x^{2}}\bigg)^{2}$

$=(x^{3})^{2}-2(x^{3})\bigg(\dfrac{1}{x^{2}}\bigg)+\bigg(\dfrac{1}{x^{2}}\bigg)^{2}$ ................. Using,

$(a-b)^2= a^2-2ab+b^2$

$=x^{6}-2{x}+\dfrac{1}{x^{4}}$

Sum of coefficients of

$x^{6}$ is

$1$ .

Find the sum $\sum _{ r=1 }^{ n }{ \overset { n+r }{ \underset { \quad \quad r }{ C }}}$ .

$\sum _{r=0} ^n\space \space ^{n+r}C_r=\space ^n C_0+ \space ^{n+1}C_1+ ^{n+2}C_2+...^{n+n}C_n$

$\Rightarrow ^nC_n+^{n+1}C_n+^{n+2}Cn+...^{2n}C_n$

$\Rightarrow$ Coefficient of

$x^n$ in

$[(1+x)^n+(1+x)^{n+1}+....+(1+x)^{2n}]$

$\Rightarrow$ Coefficient of

$x^n$ in

$\dfrac{(1+x)^n[1-(1+x)^{n+1}]}{1-(1+x)}$

$\Rightarrow$ Coefficient of

$x^{n+1}$ in

$[(1+x)^{2n+1}-(1+x)^n]$

$\Rightarrow ^{2n+1}C_{n+1}$

$\Rightarrow \sum _{r=0} ^n\space \space ^{n+r}C_r=^{2n+1}C_{n+1}$

$\Rightarrow \space ^n C_0+\sum _{r=1} ^n\space \space ^{n+r}C_r=^{2n+1}C_{n+1}$

$\Rightarrow \space 1+\sum _{r=1} ^n\space \space ^{n+r}C_r=^{2n+1}C_{n+1}$

Therefore,

$\Rightarrow \space \sum _{r=1} ^n\space \space ^{n+r}C_r=^{2n+1}C_{n+1}-1$

Find the sum of $\sum _{ 0\le i\le j\le n } \sum { j\overset { n }{ \underset { \quad i }{ C } } }$

$\sum_{0\leq i\leq j\leq n}\sum j \space\space ^nC_i$

$\Rightarrow \sum _{r=0} ^{n-1}\space \space ^nC_r[(r+1)+(r+2)+....+(n)]$

$\Rightarrow \sum_{r=0} ^n \space \space ^nC_r[\dfrac{n+1}{2}(n-r)-\dfrac{r(n-r)}{2}]$

$\Rightarrow \dfrac{n+1}{2} \sum _{r=0} ^n\space (n-r)\space \space ^nC_r -\dfrac{n}{2}\sum_{r=0} ^n \space r \space \space ^nC_r +\dfrac{1}{2}\sum_{r=0} ^n \space r^2\space \space ^nC_r$

$\Rightarrow \dfrac{n+1}{2}\sum_{r=0}^n \space r\space \space ^nC_r-\dfrac{n}{2}\sum _{r=0}^n \space r\space \space ^nC_r+\dfrac{1}{2}\sum _{r=0}^n\space r^2 \space \space ^nC_r$

$\Rightarrow \dfrac{1}{2}(\sum_{r=0} ^n\space r\space \space ^nC_r+\sum_{r=0}^n\space r^2\space \space ^nC_r)$

$\Rightarrow \dfrac{1}{2}(n 2^{n-1}+n(n-1)2^{n-2}+n2^{n-1})$

$\Rightarrow n(n+3)2^{n-3}$

Therefore,

$\Rightarrow \sum_{0\leq i\leq j\leq n}\sum j \space\space ^nC_i=n(n+3)2^{n-3}$

In the expansion of ${\left( {\dfrac{3}{2} + \dfrac{7}{3}} \right)^n}$ when $x = \dfrac{1}{2}$ if is known that ${6^{th}}$ term is the greatest term the find the possible integral value of n.

We know that ,

General term of ${{\left( x+a \right)}^{n}}$ is

${{t}_{r+1}}{{=}^{n}}{{C}_{r}}{{x}^{n-r}}.{{a}^{r}}$

Given that 6^th term is greatest in expansion of ${{\left( \dfrac{3}{2}+\dfrac{7}{3} \right)}^{n}}$

So , 6^th term is

${{t}_{6}}={{t}_{5+1}}{{=}^{n}}{{C}_{5}}{{\left( \dfrac{3}{2} \right)}^{n-5}}\times {{\left( \dfrac{7}{3} \right)}^{5}}$

Now

$n-5>0$

$n>5$

Hence value of n should be greater than 5 like 6

Hence n=6

Find the sum $\sum _{ r=1 }^{ n }{ \dfrac { r\overset { n }{ \underset { \quad \quad r }{ C } } }{ \overset { n }{ \underset { \quad \quad r-1 }{ C } } } }$

$\sum _{r=1}^n \dfrac{r\space \space ^nC_r}{^nC_{r-1}}$

$\Rightarrow \sum _{r=1}^n \dfrac{r\space \space ^nC_r}{^nC_{r-1}}=\sum_{r=1}^n r\dfrac{n-r+1}{r}$

$\Rightarrow (n+1)\sum_{r=1}^n 1-\sum_{r=1} ^n \space r$

$\Rightarrow n(n+1)-\dfrac{n(n+1)}{2}$

$\Rightarrow \dfrac{n(n+1)}{2}$

Therefore,

$\sum _{r=1}^n \dfrac{r\space \space ^nC_r}{^nC_{r-1}}=\dfrac{n(n+1)}{2}$

Find the middle term in the expansion of $\left ( \dfrac{x}{y}-\dfrac{y}{x} \right )^7$ .

${\left( \cfrac{x}{y} - \cfrac{y}{x} \right)}^{7}$

$\because n = 7 \left(\text{odd} \right)$

There will be two middle terms, i.e.,

${\left( \cfrac{n + 1}{2} \right)}^{th}$ term and

${\left( \cfrac{n + 3}{2} \right)}^{th}$ term.

$\because n = 7$

Middle term in the expansion

${\left( \cfrac{x}{y} - \cfrac{y}{x} \right)}^{7}$ will be

${4}^{th}$ and

${5}^{th}$ terms.

As we know that general term of the expansion

${\left( a + b \right)}^{n}$ will be-

${T}_{r+1} = {_{n}{C}_{r}} {a}^{n-r} {b}^{r}$

Therefore,

${4}^{th}$ term in the expansion

${\left( \cfrac{x}{y} - \cfrac{y}{x} \right)}^{7}$ -

${T}_{3+1} = {_{7}{C}_{3}} {\left( \cfrac{x}{y} \right)}^{4} {\left( \cfrac{y}{x} \right)}^{3} = {_{7}{C}_{3}} \left( \cfrac{{x}^{4}}{{y}^{4}} \right) \left( \cfrac{{y}^{3}}{{x}^{3}} \right) = {_{7}{C}_{3}} \cfrac{x}{y}$

${5}^{th}$ term in the expansion

${\left( \cfrac{x}{y} - \cfrac{y}{x} \right)}^{7}$ -

${T}_{4+1} = {_{7}{C}_{4}} {\left( \cfrac{x}{y} \right)}^{3} {\left( \cfrac{y}{x} \right)}^{4} = {_{7}{C}_{4}} \left( \cfrac{{x}^{3}}{{y}^{3}} \right) \left( \cfrac{{y}^{4}}{{x}^{4}} \right) = {_{7}{C}_{4}} \cfrac{y}{x}$

Hence the middle term of the expansion

${\left( \cfrac{x}{y} - \cfrac{y}{x} \right)}^{7}$ are-

${T}_{4} = {_{7}{C}_{3}} \cfrac{x}{y}$ and

${T}_{5} = {_{7}{C}_{4}} \cfrac{y}{x}$

Write the following form in expanded form:-
(A) ${\left( {2a + 3b} \right)^3}$
(B) ${\left( {5x - 3y} \right)^3}$

(A)

$=(2a+3b)^3$

$=$

${ C }_{ 0 }^{ 3 }{ (2a) }^{ 3 }+{ C }_{ 1 }^{ 3 }{ (2a) }^{ 2 }{ (3b) }^{ 1 }+{ C }_{ 2 }^{ 3 }{ (2a) }^{ 1 }{ (3b) }^{ 2 }+{ C }_{ 3 }^{ 3 }{ (3b) }^{ 3 }$

$=$

${ (2a) }^{ 3 }+3{ (2a) }^{ 2 }{ (3b) }^{ 1 }+3{ (2a) }^{ 1 }{ (3b) }^{ 2 }+{ (3b) }^{ 3 }$

$=$

${ 8a }^{ 3 }+36{ a }^{ 2 }b+54{ a }{ b }^{ 2 }+27{ b }^{ 3 }$

(B)

$=(5x-3y)^3$

$=$

${ C }_{ 0 }^{ 3 }{ (5x) }^{ 3 }+{ C }_{ 1 }^{ 3 }{ (5x) }^{ 2 }{ (-3y) }^{ 1 }+{ C }_{ 2 }^{ 3 }{ (5x) }^{ 1 }{ (-3y) }^{ 2 }+{ C }_{ 3 }^{ 3 }{ (-3y) }^{ 3 }$

$=$

${ (5x) }^{ 3 }+3{ (5x) }^{ 2 }{ (-3y) }^{ 1 }+3{ (5x) }^{ 1 }{ (-3y) }^{ 2 }+{ (-3y) }^{ 3 }$

$=$

${ 125x }^{ 3 }-225{ x }^{ 2 }y+135{ x }{ y }^{ 2 }-27{ y }^{ 3 }$

Show that the middle term in the expansion of ${(1 + x)^{2n}}$ is $\frac{{1.3.5.....(2n - 1)}}{{n!}}$ ${2^n}{x^n}$ ; where n is a positive integer.

In the expansion of

${\left( {1 + x} \right)^{2n}}$ in total 2n+1 terms are present.

So, middle term will be,

$\begin{array}{l}\frac{{\left( {2n + 1} \right) + 1}}{2} = \frac{{2n + 2}}{2}\\ = n + 1\end{array}$

So, middle term will be (n+1)th term.

So,

$\begin{array}{c}{a_n}{ = ^{2n}}{C_n}{\left( 1 \right)^n}{\left( x \right)^{2n - n}}\\ = \frac{{2n!}}{{n!\left( {2n - n} \right)!}}{x^n}\\ = \frac{{{2^n}\left( {n!} \right)1.3.5....\left( {2n - 1} \right)}}{{n!n!}}{x^n}\\ = \frac{{1.3.5....\left( {2n - 1} \right)}}{{n!}}{2^n}{x^n}\end{array}$

Hence proved.

If sum of coefficient in expansion of $(2+3cx+c^2x^2)^{12}$ is $0$ . Then find the value of $c$ .

For sum of coefficient in

$(2+3cx+c^{2}x^{2})^{12}$

$\Rightarrow (C^{2}+3c+2)^{12}=0$

$\Rightarrow (C+1)(C+2)=0$

$C=-1, -2$

Evaluate:
$\sum\limits_{r=0}^{20}\ ^{20}C_r$ .

We've,

$(1+x)^n=\ ^nC_0+^nC_1x+......+^nC_nx^n$ .

Now putting

$n=20$ and

$x=1$ we get,

$2^{20}=\sum\limits_{r=0}^{20}\ ^{20}C_r$ .

$^{14}C_1+$ $^{14}C_2+$ $^{14}C_3+$ ... $+$ $^{14}C_{14}=?$

1328614_1083940_ans_20cec0a5ee224d629ca697bd72a3bd8e.jpg

Is $x^5$ in the expansion of $\left[2x^3 - \dfrac{1}{3x^3}\right]^{10}$ .

$\left[2x^{3}-\dfrac{1}{3x^{3}}\right]^{10}$

$T_{r+1}=\ ^{10}C_{r}(2x^{3})^{10-r}\left(\dfrac{-1}{3x^{3}}\right)^{r}$

$=\ ^{10}C_{r}2^{10-r}.x^{30-3r-3r}\left(-\dfrac{1}{3}\right)^{r}$

$=\ ^{10}C_{r}.2^{10-r}.x^{30-6r}\left(-\dfrac{1}{3}\right)^{r}$

We need

$x^{5}$ term so put

$30-6r=5$

$6r=25$

$r=\dfrac{25}{r}$

it cant be fraction

So,

$x^{5}$ is not define in the expansion.

Find the $24th$ term in the sequence whose nth term ${a_n} = \dfrac{{n\left( {n - 2} \right)}}{{n + 3}}$ . using binomial theorem , Evaluate ${\left( {102} \right)^3}$

$\\a_{24}=(\dfrac{24(24-2)}{24+3})$

$\\=(\dfrac{24 \cdot 22}{27})$

$\\=(\dfrac{176}{9})$

$\\Now\>(102)^3=(100+2)^3$

$\\={^3}C_0(100)^3 \cdot 2^0 + {^3}C_1(100)^2 \cdot2^1 + {^3}C_2(100)^1 \cdot 2^2+{^3}C_3 (100)^0 \cdot 2^3$

$=1000000+60000+1200+8$

$=1061208$

Find the $7$ th term in the expansion of $\left(\dfrac{4x}{5}-\dfrac{5}{2x}\right)^9$ .

The binominal expansion of

$\left(\dfrac{4x}{5}-\dfrac{5}{2x}\right)^{9}$

$= \,{^{a}C_{0}} \left(\dfrac{-5}{2x}\right)^{0} \left(\dfrac{4x}{5}\right)^{9-0}+^{9}C_{1}\left(\dfrac{-5}{2x}\right)^{1} \left(\dfrac{4x}{5}\right)^{9-1}+.....+^{a}C_{9} \left(\dfrac{-5}{2x}\right)^{9} \left(\dfrac{4x}{5}\right)^{9-9}$

$=\displaystyle \sum_{r=0}^{9} {^{a}C_{r}} \left(\dfrac{-5}{2x}\right)^{r} \left(\frac{4x}{5}\right)^{9-r}$

the

$7^{th}$ term would br for

$r=6$

i.e,

$^{9}C_{6} \left(\dfrac{-5}{2x}\right)^{6} \left(\dfrac{4x}{5}\right)^{3}$

$=84.\dfrac{5^{6}}{2^{6}.x^{6}}.\dfrac{4^{3}.x^{3}}{5^{3}}$

$=84.5^{3}x^{-3}$

Find the coefficient of $x^4$ in $(1 + 2x)^4 (2 -x)^5$

1097441_1137948_ans_e900813f5baa49c398fe80d7ae166992.jpg

Find the middle terms in the expansions of
i) $\left(3 - \dfrac{x^3}{6} \right)^7$
ii) $\left(\dfrac{x}{3} + 9y \right)^{10}$

$i)$

$\left ( 3-\dfrac{x^3}{6} \right )^7$

$n=7\Rightarrow$ middle terms are

$T_4,T_5$

$T_{r+1}=^nC_ra^{n-r}b^r$

$T_4=T_{3+1}=^7C_33^{7-3}\left ( -\dfrac{x^3}{6} \right )^3$

$=-\dfrac{7\times 6\times 5}{3\times 2}\cdot 3\cdot \dfrac{x^9}{8}$

$\therefore T_4=-\dfrac{105}{8}x^9$

$T_5=T_{4+1}=^7C_43^{7-4}\left ( -\dfrac{x^3}{6} \right )^4$

$=-\dfrac{7\times 6\times 5}{3\times 2}\cdot 1\cdot \dfrac{x^{12}}{16}$

$\therefore T_5=\dfrac{35}{48}x^{12}$

$ii)$

$\left ( \dfrac{x}{3} +9y\right )^{10}$

$n=10\Rightarrow$ middle term

$=\dfrac{n}{2}+1=6$

$T_{r+1}=^nC_ra^{n-r}b^r$

$T_{5+1}=^{10}C_5\left (\dfrac{x}{3} \right )^{10-5}(9y)^5$

$=\dfrac{10\times 9\times 8\times 7\times 6\times 3^5}{5\times 4\times 3\times 2\times 1}\cdot x^5\cdot y^5$

$T_6=61236x^5y^5$

Find the coeifficient of
(i) $x^{6}y^{3}in (x+y)^{9}$

$x^6 y^3$ in

$(x + y)^9$

We know general term of expansion

$(a + b)^n$ is

$T_{r + 1} = \, ^nC_r a^{n- r} b^r$

For

$(x + y)^9$

Putting

$n = 9 , a = x, b = y$

$T_{r + 1} = \, ^9C_r (x)^{9 - r} (y)^r$ ___(i)

we need to find coefficient of

$x^6 y^3$

Comparing,

$y^r = y^3$

$r = 3$

Putting

$r = 3$ in (i)

$T_{3 + 1} = ^9C_3 (x)^{9 - 3} (y)^3$

$= \dfrac{9!}{3! 6!} (x)^6 (y)^3$

$= \dfrac{9 \times 7 \times 6!}{3 \times 2 \times 6!} (x)^6 (y)^3$

$= \dfrac{3 \times 7}{2} (x)^6 (y)^3$

$= \dfrac{21}{2} (x)^6 (y)^3$

Hence, coefficient of

$x^6 y^3$ is

$\dfrac{21}{2}$

Show that the middle term in the expansion of $(1+x)^{2n}$ is $\dfrac{1.3.5...(2n-1)}{n!}$ $2^nx^n$ , where n is a positive integer.

$(1+x)^{2n} = ^{2n}C_{0}x^{0}+^{2n}C_{1}x^{1}+---+^{2n}C_{2n}x^{2n}$

$\displaystyle = \sum_{r = 0}^{2n} x^{r}. ^{2n}C_{r}$

So, there are (2n+1) terms in the expansion

the middle term be for r=n

which is

$^{2n}C_{n}.x^{n}$

$= \dfrac{(2n)!}{n!(2n-n)!}x^{n}$

$= \dfrac{2n.(2n-1)(2n-2)---4.3.2.1}{n!. n!}x^{n}$

$= \dfrac{(1.3.5---(2n-1))2^{n}(1.2.3...n)}{n! \, n!} ,x^{n}.$

$= \dfrac{1.3.5---(2n-1)}{n!}2^{n}x^{n}.$

Find the coefficient of $x^{15}$ in
$(1 + x)^{15} + (1 + x)^{16} + .... (1 + x)^{30}$ .

1363985_1133416_ans_41483646491a437f988446c1b64ac328.jpg

If the coefficients of $\left( r-5 \right)$ th and $\left( 2r-1 \right)$ th terms in the expansion of ${ \left( 1+x \right) }^{ 34 }$ are equal, then write the value of $r$

Coefficient of any tern in expansion

$(1+x)^{34}$ is given as :-

$^{34}C_r\,\, 1^{34-r}x^r$

$=^{34}C_rx^r$

$\therefore T_{r+1}=^{34}C_rx^r$

$\therefore (T_{r-6+1})$ coefficient =

$(T_{2r-2+1})$ coefficient

$\therefore ^{34}C_{r-6}=^{34}C_{2r-2+1)}$

Only possible when

$r-6=2r-2$ OR

$r-6+2r-2=34$

$\therefore r=-4$ OR

$3r=42$

$r$ can't be negative

and

$\therefore r=\dfrac{42}{3}=14$

$\therefore$ Value of

$r$ is

$\boxed{14}$

Find the $8th$ term in the expansion of ${ \left( { x }^{ 3/2 }{ y }^{ 1/2 }-{ x }^{ 1/2 }{ y }^{ 3/2 } \right) }^{ 10 }$

$(a+b)^{n} \Rightarrow$ General term

$\Rightarrow T_{r} = ^{n}C_{r} a^{n-r} b^{r}$

$\therefore 8^{th}$ term in

$(x^{3/2} y^{1/2}-x^{1/2}y^{3/2})^{10}$

$T_8 \Rightarrow ^{10}c_{7}(x^{3/2}y^{1/2})^{1/2}(-1)(x^{1/2}y^{3/2})^{7}$

$= \dfrac{10 \times 8 \times 9 (-1)}{3 \times 2} (x^{1/2}y^{3/2}) (x^{3/2} y^{2/2} )$

$= -10 \times 4 \times 3 (x^{9/2+7/2}) (y^{3/2+4/2})$

$= -120 (x^{16/2}y^{24/2})$

$T_{8} = -120x^{8}y^{12}$

1116232_1139899_ans_279f3fb125ec482cb868f90075af8aea.jpeg

Prove that there is no term involving $x^6$ in the expansion of $\left (2x^2-\dfrac{3}{x}\right)^{11}$ , where $r\neq 0$ .

Suppose

$x^6$ occurs in

$(r+1)^{th}$ term in the expansion of

$(2x^2-\dfrac{3}{x})^{11}$ .

Now,

$T_{r+1}=11_{C_r}(2x^2)^{11-r} (-\dfrac{3}{x})^r$

$=11_{C_r} (-1)^r 2^{11-r} 3^r x^{22-3r}$

For this term to contain

$x^6$ ,

$22-3r=6$

$r=\dfrac{16}{3}$ , which is a fraction.

But, r is a natural number. Hence, there is no term containing

$x^6$ .

If the coefficients of $x$ and ${x}^{2}$ in the expansion of ${(1+x)}^{m}{(1-x)}^{n}$ are $3$ and $-6$ respectively. Find the values of $m$ and $n$ .

We have

$(1+x)^m(1-x)^n$

$=[1+^mC_1x+^mC_2x^2+^mC_3x^3+.....][1-^nC_1x+^nC_2x^2-^nC_3x^3+....]$

Coefficient of x will be obtained when a constant term in the first bracket is multiplied with term having x in second bracket

Given that coefficient of

$x=3$

$\Rightarrow ^mC_1-^nC_1=3$

$\Rightarrow m-n=3$

$\Rightarrow m=n+3.....(1)$

Coefficient of

$x^2=-6$

$\Rightarrow ^mC_2+^nC_2-^mC_1.^nC_1=-6$

$\Rightarrow \dfrac{m(m-1)}{2}+\dfrac{n(n-1)}{2}-mn=-6$

$\Rightarrow \dfrac{(n+3)(n+3-1)}{2}+\dfrac{n(n-1)}{2}-(n+3)n=-6$

$\Rightarrow (n+3)(n+2)+n(n-1)-2(n+3)n=-12$

$\Rightarrow n^2+5n+6+n^2-n-2n^2-6n+12=0$

$\Rightarrow 2n^2-2n^2-2n+18=0$

$\Rightarrow 2n=18$

$\Rightarrow n=9$

Putting in (1)

$m=n+3$

$m=9+3$

$m=12$

So the value of m and n are 12 and 9

Find the coefficient of x $^{13}$ in the expansion of $(1-x)^{5}x(1+x+x^{2}+x^{3})^{4}$

$(1-x)^5x(1+x+x^2+x^3)^4$ sum of

$4$ terms of G.P

$=1+x+x^2+x^3=1\dfrac{(1-x^4)}{1-x}$

$=(1-x)^5x\left(\dfrac{1-x^4}{1-x}\right)^4$

$=(1-x)^5x\dfrac{(1-x^4)^4}{(1-x)^4}$

$=(1-x)x(1-x^4)^4$

Coefficient of

$x^{13}$ in

$(1-x)x(1-x^4)^4$

$=$ Coefficient of

$x^{12}$ in

$(1-x)(1-x^4)^4$

$=$ Coefficient of

$x^{12}$ in

$(1-x^4)^4-x(1-x^4)^4$

$=$ Coefficient of

$x^{12}$ in

$(1-x^4)^4-$ coefficient of

$x^{12}$ in

$x(1-x^4)^4$

$=$ Coefficient of

$x^{12}$ in

$(1-x^4)^4-0$ (

$x^{12}$ term is not possible in

$x(1-x^4)^4$ )

$={^{4}C_3}(-1)^3$

$(\because (1-x^4)^4={^{4}C_0}-{^{4}C_1}x^4+{^{4}C_2}x^8-{^{4}C_3}x^{12}+{^{4}C_4}x^{16})$

$=-4$

$\therefore$ Coefficient of

$(1-x)^5x(1+x+x^2+x^3)^4=-4$ .

1193047_1195345_ans_adde32a60f2342f1ba9fd5b94a052bc7.jpg

The sum of the coefficients of first three terms in the expansion of ${ \left( x-\cfrac { 3 }{ { x }^{ 2 } } \right) }^{ m },x\neq 0$ , $m$ being a natural number, is $559$ . Find the term of the expansion containing ${x}^{3}$ .

Given,

$\left(x - \dfrac{3}{x^2} \right)^m$

$T_{r + 1} = ^m C_r x^{m -r} \, \left(-\dfrac{3}{x^2} \right)^r$

$T_{r + 1} = ^m C_r (-3)^r . x^{m - 3r}$

Given,

$^mc_0 (-3)^0 + ^mC_1 (-3)^1 + ^mC_2 (-3)^2 = 559$

$\Rightarrow 1 + m (-3) + 9 \times \dfrac{m (m - 1)}{2} = 559$

$\Rightarrow 2 - 6m + 9m^2 - 9m = 559 \times 2$

$\Rightarrow 9m^2 - 15 m - 1116 = 0$

$m = 12, m = \dfrac{-31}{3} (Impossible)$

$\therefore m = 12$

$Tr + 1 = 12 C_r (-3)^r x^{12 - 3r}$

$12 - 3r = 3$

$\therefore r = 3$

$T_3 = 12 C_2 (-3)^2$

$T_3 = 594$

If ${x}^{p}$ occurs in the expansion of $\quad { \left( { x }^{ 2 }+\cfrac { 1 }{ x } \right) }^{ 2n }$ , prove that its coefficient is $\quad \left[ \cfrac { \left( 2n \right) ! }{ \left( \cfrac { 4n-p }{ 3 } \right) !\left( \cfrac { 2n+p }{ 3 } \right) ! } \right]$

$\left(x^2 + \dfrac{1}{x}\right)^{2n}$

$T_{r + 1} = ^{2n} Cr (x^2)^{2n - r} \left(\dfrac{1}{x}\right)^r$

$T_{r +1} = ^{2n} Cr \, x^{4n - 2r - r}$

$T_{r + 1} = ^{2n} Cr x^{4n - 3r}$

$4n - 3r = p$

$\Rightarrow 4n - p = 3r$

$\Rightarrow r = \dfrac{4n - p}{3}$

$\therefore$ coefficient of

$x^p$ is

$^{2n}C_{\dfrac{4n - p}{3}}$

$= \dfrac{(2n)!}{\left(\dfrac{4n - p}{3} \right) ! \left(\dfrac{2n + p}{3} \right)!}$

The value of $\sum_{r=1}^{15}r^{2}(\frac{^{15}Cr}{^{15}Cr-1})$ is equal to:

$\displaystyle\sum^{15}_{r=1}r^2\left(\dfrac{^{15}C_r}{^{15}C_{r-1}}\right)$

$\dfrac{^{n}C_r}{^{n}C_{r-1}}=\dfrac{n-r+1}{r}$

$\dfrac{^{15}C_r}{^{15}C_{r-1}}=\dfrac{15-r+1}{r}=\dfrac{16-r}{r}$

$\displaystyle\sum^{15}_{r=1}r^2\left(\dfrac{^{15}C_r}{^{15}C_{r-1}}\right)=\displaystyle\sum^{15}_{r=1}r^2\left(\dfrac{16-r}{r}\right)$

$=\displaystyle\sum^{15}_{r=1}16r-r^2$

$=16\displaystyle\sum^{15}_{r=1}r-\displaystyle\sum^{15}_{r=1}r^2$

$\displaystyle\sum^n_{r=1}r=\dfrac{n(n+1)}{2}$

$\displaystyle\sum^n_{r=1}r^2=\dfrac{n(n+1)(2n+1)}{6}$

$=16\left(\dfrac{15(15+1)}{2}\right)-\dfrac{15(15+1)(2(15)+1)}{6}$

$=15(16)\left[8-\dfrac{31}{6}\right]=15(16)\dfrac{17}{6}$

$=5\times 8\times 17=40\times 17$

$=680$ .

1180065_1194615_ans_e5897b5f039a441dbde4c817e2a792eb.jpg

The coefficient of x $^{10}$ in the expansion of (1+x) $^{2}$

$(1+x)^2 = x^2+2*x+1$

Co-efficient of

$x^10$ in expansion is "0".

If the coefficients of $r$ th, $(r+1)$ th and $(r+2)$ nd terms in the expansion of ${(1+x)}^{n}$ are in A.P, then show that ${n}^{2}-(4r+1)n+4{r}^{2}-2=0$ .

${(1+x)}^{n}$

${T}_{r+1}={ _{ }^{ n }{ C } }_{ r }{x}^{r}$

$2{T}_{r+1}={T}_{r}+{T}_{r+2}$

$\Rightarrow$

$2{ _{ }^{ n }{ C } }_{ r }={ _{ }^{ n }{ C } }_{ r-1 }+{ _{ }^{ n }{ C } }_{ r+1 }$

$\Rightarrow$

${n}^{2}-(4r+1)n+4{r}^{2}-2=0$

If $\displaystyle \sum_{r = 1}^{n} {^{n}C_{r}} 3^{r}$ is equal to $4095$ then $n$ equals.

We have,

$^n_{r=1}\sum ^nC_r=4095$

$\implies ^nC_1+^nC_2 3^2+....^nC_n 3^n=4095$

$\implies (1+3)^n-^nC_0=4095$

$\implies 4^n-1=4095$

$\implies 4^n=4^6$

$\implies n=6$

Hence, this is the answer.

The coefficent of $x^{10}$ in the expansion $(x+10)^{10}$

Let

$Tr+1={ n }_{ { C }_{ r } }{ x }^{ n-r }x{ \left( 10 \right) }^{ r }$

Now

$n=10$ and

$n-r=10$

so,

$r=0$

$Tr+1={ 10 }_{ { C }_{ 0 } }{ x }^{ 10 }{ \left( 10 \right) }^{ 0 }={ T }_{ 1 }$

So, coefficient

$={ 10 }_{ { C }_{ 0 } }\times { \left( 10 \right) }^{ 0 }=1$

Find the middle term in the expansion of $\left (\dfrac{2}{3}x^2-\dfrac{3}{2x}\right )^{20}$ .

1341462_1197413_ans_085d78e610464f18b95ebd2c760bfb62.jpg

Find the middle term of $\left(x-\dfrac {1}{2x}\right)^{10}$

$S=\left( x-\dfrac { 1 }{ 2x } \right) ^{ 10 }$

General term of a binomial expansion,

$(x+y)^{ n }=\sum _{ r=0 }^{ n }{ ^nC_rx^ry^{n-r} }$

Here,

$n=10$ so there are 11 terms, middle term is the sixth term,

$r=5$

$T_6=^{10}C_5x^5(-1/2x)^5=-^{10}C_52^{-5}$ .

Find the middle term in term expansion of ${\left(1+x\right)}^{2n}$

1339873_1232512_ans_74eba22708984646b5b49c8cb2ad2548.PNG

If the sum of the coefficients of $x^7 and \ x^4$ in the expansion of $\left(\dfrac{x^2}{a} - \dfrac{b}{x}\right)^{11}$ is zero, then ab=

General term,

$T_{r + 1} = ^{11}C_{r} \left (\dfrac {x^{2}}{a}\right )^{r} \left (\dfrac {-b}{x}\right )^{11 - r}$

$= ^{11}C_{r} a^{-r} (-b)^{11 - r} . x^{3r -11}$

$coeff. \, of \, \underline {x^{7}} \quad \ = ^{11}C_{7} a^{-7} (-b)^{4}$ ..............(i)

$coeff. \, of \, \underline {x^{4}} \quad =^{11}C_{4} a^{-4} (-b)^{7}=- ^{11}C_{4} a^{-4} (b)^{7}$ ................(ii)

(i)

$+$ (ii)

$=0$

$^{11}C_{7} a^{-7} (-b)^{4}- ^{11}C_{4} a^{-4} (b)^{7}=0$

$\Rightarrow ^{11}C_{7} a^{-7} (-b)^{4} =^{11}C_{4} a^{-4} (b)^{7}$

$^{11}C_{4} = ^{11}C_{7} \Rightarrow \dfrac {b^{4}}{a^{7}} = \dfrac {b^{7}}{a^{4}}$

$\Rightarrow a^{3}b^{3} = 1\Rightarrow ab = 1$ .

Find general term in the expansion of $\left( x _ { 1 } + x _ { 2 } + \ldots . + x _ { p } \right) ^ { n }$

the general term in the expansion of

$(x_1+x_2+\cdots+x_p)^{n}$ is

$\dfrac{n!}{n_1! n_2! \cdots n_p !}x_1^{n_1}x_2^{n_2}\cdots x_p^{n_p}$

In the expansion of $\left( 5 ^ { \dfrac { 1 } { 2 } } + 2 ^ { \dfrac { 1 } { 8 } } \right) ^ { 1024 },$ the number of integral terms is

${ T }_{ r+1 }={ 1024 }_{ Cr }{ \left( { 5 }^{ 1/2 } \right) }^{ 1024-r }.{ \left( { 7 }^{ 1/8 } \right) }^{ r }$

for

${ T }_{ r+1 }$ to be integer,

$\dfrac { 1024-r }{ 2 } \& \dfrac { r }{ 8 }$ should be integer

$r/8$ is integer,

$r\le 1024$

$\Rightarrow \dfrac { r }{ 8 } \le \dfrac { 1024 }{ 8 }$

$\Rightarrow \dfrac { r }{ 8 } \le 128$

$\therefore$ No. of terms

$=128+1=129$

Find the coefficient of ${ x }^{ 3 }$ in the expansion of ${ \left( { x }^{ 2 }+\dfrac { 3\sqrt { 2 } }{ x } \right) }^{ 9 }$

using binomial therm

$(x^{2}+\frac{3\sqrt{2}}{x^{9}})$

$\sum_{n=0}^{9}$

$^{9}C_{n}$

$(x^{2})^{9-n}(\frac{3\sqrt{2}}{x})^{n}$

$\sum_{n=0}^{9}$

$^{9}C_{n}$

$x^{18-2n}$

$\frac{(3\sqrt{2})^{n}}{x^{n}}$

$= \sum_{n=0}^{9}$

$^{9}C_{n}$

$x^{18-3n}$

$(3\sqrt{2})^{n}$

$=\sum_{n=0}^{9}$

$^{9}C_{n}$

$x^{18-2n}$

$T_{n+1} = ^{9}C_{n} (3\sqrt{2})^{n} x^{18-3n}$

coefficient of

$x^{3} = ^{9}C_{5} (3\sqrt{2})^{5} = ^{9}C_{5} 3^{5}. (\sqrt{2})^{5} [18+3n = 3 3n=18-3 n =\frac{15}{3}=5$

$= \frac{9\times 8\times 7\times 6\times 5}{1\times 2\times 3\times 4\times 5} 3^{5}.4\sqrt{2}$

$= 126. 243 . 4\sqrt{2} .$

$= 422,472\sqrt{2}.$

1190611_1331492_ans_e60b7c2b31954dbe81a297aca26b2dfa.JPG

Find the coefficient ${ x }^{ 8 }$ in the product $\left( 1+2x \right) ^{ 6 }\left( 1-x \right) ^{ 7 }$ using binomial therm.

Given,

$(1+2x)^6(1-x)^7$

$(1+2x)^6=^6C_0+^6C_1(2x)+^6C_2(2x)^2+^6C_3(2x)^3+^6C_4(2x)^4+^6C_5(2x)^5+^6C_6(2x)^6$

$=1+12x+60x^2+160x^3+240x^4+192x^5+64x^6$ ......(1)

$(1-x)^7=^7C_0-^7C_1(x)+^7C_2(x)^2-^7C_3(x)^3+^7C_4(x)^4-^7C_5(x)^5+^7C_6(x)^6-^7C_7(x)^7$

$=1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7$ .........(2)

multiplying (1) and (2), we get,

$(1+2x)^6(1-x)^7=(1+12x+60x^2+160x^3+240x^4+192x^5+64x^6)(1-7x+21x^2-35x^3+35x^4-21x^5+7x^6-x^7)$

$x^8$ coefficients are,

$=(12x)(-x^7)+(60x^2)(7x^6)+(160x^3)(-21x^5)+(240x^4)(35x^4)+(192x^5)(-35x^3)+(64x^6)(21x^2)$

$=-12x^8+420x^8-3360x^8+8400x^8-6720x^8+1344x^8$

$=72x^8$

coefficeint of

$x^8=72$

If the middle term in the expansion of $\left(x+\dfrac {b}{x}\right)^{6}$ is $160$ , find $b$ .

$\left(x+\dfrac{b}{x}\right)^{6}$ Total term

$=7$

middle term=

$\dfrac{7+1}{2}=\dfrac{8}{2}=4^{th}$

$T_{4}$

$= ^{6}C_{3}$

$x^{3}$

$\left(\dfrac{b}{x}\right)^{3}$

$160=\dfrac{6!}{3!3!}.x^{3}\dfrac{b^{3}}{x^{3}}$

$b^{3}=8$

$\Rightarrow b=2$ .

1215250_1353771_ans_9f884754ec074d5790f26185fae8367f.jpg

Find the sum of coefficients of odd powers of $x$ in the expansion ${\left(1+x\right)}^{50}$

${\left(1+x\right)}^{n}=^{n}{{C}_{0}}+^{n}{{C}_{1}}x+^{n}{{C}_{2}}{x}^{2}+...+^{n}{{C}_{0}}{x}^{n}$ .... $(1)$

${\left(1-x\right)}^{n}=^{n}{{C}_{0}}-^{n}{{C}_{1}}x+^{n}{{C}_{2}}{x}^{2}-...+^{n}{{C}_{0}}{x}^{n}$ .... $(2)$

$\Rightarrow {\left(1+x\right)}^{n}-{\left(1-x\right)}^{n}=^{n}{{C}_{0}}+^{n}{{C}_{1}}x+^{n}{{C}_{2}}{x}^{2}+...+^{n}{{C}_{0}}{x}^{n}-^{n}{{C}_{0}}+^{n}{{C}_{1}}x-^{n}{{C}_{2}}{x}^{2}+...+^{n}{{C}_{0}}{x}^{n}$

$=2\left(^{n}{{C}_{1}}x+^{n}{{C}_{3}}{x}^{3}+....+{x}^{n}\right)$

Given: $n=50$ and put $x=1$

$\Rightarrow {\left(1+1\right)}^{50}-{\left(1-1\right)}^{50}={2}^{50}$

$\Rightarrow {2}^{50}=2\left(^{n}{{C}_{1}}x+^{n}{{C}_{3}}{x}^{3}+....+{x}^{n}\right)$

$\therefore$ Sum of odd powers of $x$ in the expansion is $= ^{n}{{C}_{1}}x+^{n}{{C}_{3}}{x}^{3}+....+{x}^{n}={2}^{50-1}={2}^{49}$

The coefficient of $x ^ { 10 }$ in the expansion of $( 1 + x ) ^ { 2 } \left( 1 + x ^ { 2 } \right) ^ { 3 } \left( 1 + x ^ { 3 } \right) ^ { 4 }$ is equal to :

$k=4, m=0, n=3$

$\therefore$ coefficient of

$x^{10}$ is:

$^9C_4 \times ^3C_2 \times ^4C_3$

$=\dfrac{9!}{5! 4!} \times \dfrac{3 !}{!!2!} \times \dfrac{4 !}{!! 3!}$

$=\dfrac{9 \times 12 \times 6 \times 5!}{5! \times 3} \times \dfrac{3 \times 2!~}{2!} \times \dfrac{4 \times 3!}{2!}$

$=3 \times 7 \times 6 \times 3 \times 4$

$=1512$

NOw,

$x^{9-k}.(x^2)^{3-m}.(x^3)^{4-n}=x^{10}$

$\Rightarrow x^{5-k+6-2m+12-3n}=x^{10}$

$\Rightarrow x^{27-k-2m-3n}=x^{10}$

$\therefore 27-k-2m-3n=10$

Now,

$k \le 9, m \le 3, n \le 4$

$k+2m+3m= 17$

$(1+x)^2(1+x^2)^3(1+x^3)^4$

$\left(^9C_k x^{9-k}\right) \left(^3C_m (x^n)^{3-m}\right) \left(^4C_n. (x^3)^{4-n}\right)$

Now,

$x^{9-k}.(x^2)^{3-m}.*(x^3)^{4-n}=x^{19}$

$\Rightarrow x^{9-k+6-2m+12-3n}=x^{10}$

$\Rightarrow x^{27-k-2m-3m}=x^{10}$

$\therefore 27-k-2m-3n=10$

The middle term in the expansion of $(5x-7y)^{7}$ .

Since n is an odd number, the expansion contains two middle terms.

$(\dfrac{7+1}{2})th$ and

$(\dfrac{7+3}{2})th$ terms are the two middle terms in the expansion of

$(5x-7y)^{7}$ .

$T_{4}=T_{3+1}=(-1)^{3}{7_{C}}_{3}(5x)^{7-3}(7y)^{3} = -{7_{C}}_{3} (5x)^{4}(7y)^{3}$

$T_{5}=T{_4+1}=(-1)^{4}. {7_{C}}_{4}(5x)^{7-4}(7y)^{4}={7_{C}}_{4}(5x)^{3}(7y)^{4}$ .

Fidn the ${ 7 }^{ th }$ term from the end in the expansion of ${ \left( 9x-\dfrac { 1 }{ 3\sqrt { x } } \right) }^{ 18 },x\neq 0$ .

$\left(9x-\dfrac{1}{3\sqrt{x}}\right)^{18}=7^{th}$ term from end

$11^{th}$ term from start

$T_{11}={^{18}C_{11}(9x)^{11}}\left(\dfrac{1}{3\sqrt{x}}\right)^{18-11}$ .

1210841_1358489_ans_a834d044a4984284a268593c28843ed3.jpg

In expansion of $(1+x)^3$ , the second term is 240 , find $x$

$(1+x)^3=1+3x+3x^2+x^3$

The second term is

$3x=240\\x=80$

Find the middle term in the expansion of
$\left( \dfrac { 2x }{ 3 } -\dfrac { 3 }{ 2x } \right) ^{ 6 }$

Total terms

$=6+1=7$ terms

Middle terms

$=\dfrac{7+1}{2}=4^{th}$ term

$T_4={^{6}C_3}\left(\dfrac{2x}{3}\right)^3\left(\dfrac{-3}{2x}\right)^3$

$=\dfrac{6!}{3!3!}\times (-1)^3=\dfrac{-6\times 5\times 4}{3\times 2}$

$=-20$ .

1203585_1445196_ans_89ef111011864889b4770ca9da8e3537.jpg

Find the middle term in the expansion of $\left( \dfrac { { 2x }^{ 2 } }{ 3 } -\dfrac { 3 }{ 2x } \right) ^{ 12 }$ .

$(\dfrac{2x^{2}}{3}-\dfrac{3}{2x})^{12}$

middle term be r=7

$^{12}C_{7}(\dfrac{2x^{2}}{3})^{5}.(\dfrac{-3}{2x})^{7}$

$-^{12}C_{7} 2^{5}. \dfrac{x^{10}}{3^{5}}\dfrac{3^{7}}{2^{7}.x^{7}}$

$-^{12}C_{7}\dfrac{3^{2}}{2^{2}}.x^{3}$

$-\dfrac{12!}{5!7!}\dfrac{9}{4}.x^{3}$

$\dfrac{12\times 11\times 10\times 9\times 8\times9}{5\times 4\times 3\times 2\times4}.x^{3}$

$-1782 x^{3}$

Find the 8th term of ${\left( {1 - \frac{{5x}}{2}} \right)^{ - 3/5}}$

${\left( 1 - \cfrac{5x}{2} \right)}^{- \frac{3}{5}}$

General term of the expansion

${\left( 1 + X \right)}^{n}$ is given as-

${T}_{r + 1} = \cfrac{n \left( n - 1 \right) \left( n - 2 \right) ..... \left( n - r + 1 \right)}{r!} {X}^{r}$

Now, in the given expansion, i.e.,

${\left( 1 - \cfrac{5x}{2} \right)}^{- \frac{3}{5}}$

$X = - \cfrac{5x}{2}$

$n = - \cfrac{3}{5}$

Therefore,

${T}_{8} = {T}_{7 + 1}$

$\therefore r = 7$

${T}_{8} = \cfrac{\left( - \cfrac{3}{5} \right) \left( - \cfrac{3}{5} - 1 \right) \left( - \cfrac{3}{5} - 2 \right) ..... \left( - \cfrac{3}{5} - 7 + 1 \right)}{7!} {\left( - \cfrac{5x}{2} \right)}^{7}$

$\Rightarrow {T}_{8} = - {\left( \cfrac{5}{2} \right)}^{7} \cfrac{\left( -3 \right) \left( -8 \right) \left( -13 \right) ..... \left( -33 \right)}{{5}^{7} 7!} {x}^{7}$

$\Rightarrow {T}_{8} = \cfrac{3 \cdot 8 \cdot 13 \cdot ..... \cdot 33}{{2}^{7} 7!} {x}^{7}$

Hence the

${8}^{th}$ term of the given expansion is

$\cfrac{3 \cdot 8 \cdot 13 \cdot ..... \cdot 33}{{2}^{7} 7!} {x}^{7}$ .

Find the ${ 7 }^{ th }$ term in ${ \left( \dfrac { 4 }{ { x }^{ 3 } } +\dfrac { { x }^{ 2 } }{ 2 } \right) }^{ 14 }$

$\left(\dfrac{4}{x^3}+\dfrac{x^2}{2}\right)^{14}$

$T_7={^{14}C_6}\left(\dfrac{4}{x^3}\right)^8\left(\dfrac{x^2}{2}\right)^6$

$=\dfrac{14!}{6!8!}\times \dfrac{4^8}{x^{24}}\times \dfrac{x^{12}}{2^6}$

$=\dfrac{14\times 13\times 12\times 10\times 9}{6\times 5\times 3\times 2}\times \dfrac{2^{16}}{2^6}\times \dfrac{1}{x^{12}}$

$=\dfrac{14\times 13\times 6\times 2^{10}}{x^{12}}$

$=\dfrac{1118208}{x^{12}}$ .

1209780_1459448_ans_714feae6b7be41f9b9d6e5eae5b37da0.jpg

Prove that the coefficient of ${x^n}$ in the expression of ${\left( {1 + x} \right)^{2n}}$ is twice the coefficient of ${x^n}$ in the expression of ${\left( {1 + x} \right)^{2n - 1}}$ .

We know that

General term of

$(a+b)^n$ is

$T_{r+1}={^{n}C_r}a^{n-r}, b^r$

For

$(1+x)^{2n}$

General term of

$(1+x)^{2n}$

Putting

$a=1, b=x, n=2n$

$T_{r+1}={^{2n}C_r}1^{2n-r}(x)^r$

$T_{r+1}={^{2n}C_r}x^r$ …………….

$(1)$

For coefficient of

$x^n$

Putting

$r=n$ in

$(1)$

$T_{r+1}={^{2n}C_n}x^n$

Coefficient of

$x^n={^{2n}C_n}$

For

$(1+x)^{2n-1}$

General term of

$(1+x)^{2n-1}$

Putting

$a=1, b=x, n=2n-1$

$T_{r+1}{^{2n-1}C_n}x^n$

Coefficient of

$x^n={^{2n-1}C_n}$

We have to prove

Coefficient of

$x^n$ in

$(1+x)^{2n}=2x$ , coefficient of

$x^n$ in

$(1+x)^{2n-1}$

i.e.,

${^{2n}C_n}={^{2n-1}C_n}$

L.H.S R.H.S

$^{2n}C_n$

$2\times {^{2n-1}C_n}$

$=\dfrac{2n!}{n!(2n-n)!}$

$=2\times \dfrac{(2n-1)!}{n!(2n-1-n)!}$

$=\dfrac{2n!}{n!\cdot n!}$

$=2\times \dfrac{(2n-1)!}{n!(n-1)!}$

Multiply & divide by n

$=2\times \dfrac{(2n-1)!}{n!(n-1)!}\times \dfrac{n}{n}$

$=\dfrac{2n(2n-1)!}{n!n(n-1)!}$

$=\dfrac{2n!}{n!n!}$

Hence L.H.S

$=$ R.H.S

We proved.

1221358_1451152_ans_c4e1b2a2b54c4f9991c53959a3ab5332.jpg

Find the middle term in the expansion of :
$\left(\dfrac{x}{a}-\dfrac{a}{x}\right)^{10}$

Given that to find middle term in expansion of

$\left(\dfrac{x}{a}-\dfrac{a}{x}\right)^{10}$

$n=10\Rightarrow$ even

For even values middle term is

$\left(\dfrac{n}{2}+1\right)th$

$\therefore$ middle term

$=\left(\dfrac{10}{2}+1\right)th=6^{th}$

We know that

$T_{r+1}=\ ^{n}C_{r} (a)^{n-r}(b)^{r}$

$T_{6}=T_{5+1}=\ ^{10}C_{5}\left(\dfrac{x}{a}\right)^{5}\left(\dfrac{-a}{x}\right)^{5}$

$=-\dfrac{10!}{5!5!}\dfrac{x^{5}}{a^{5}}\dfrac{a^{5}}{x^{5}}$

$=-252$

Find the coefficient of $x^3$ in the expansion of $(3x + 1)(2 - x)^6$

$(2 - x)^6=64-192x+240x^2-160x^3+20x^4-12x^5+x^6$

Coefficient of

$x^2$ is

$240$

Coefficient of

$x$ is

$3$

Coefficient of

$x^3$ is

$720-160=560$

Find the coefficient of $x^2$ and $x^3$ in the expansion of $(1-2x)^3$ .

$(1-2x)^3\\=1-3(1)^2(2x)+3(1)(2x)^2-(2x)^3\\=1-6x+12x^2-8x^3$

The coefficient of

$x^2$ is

$12$

The coefficient of

$x^3$ is

$-8$

Show that the expansion of $(2x^2 - \frac {1}{x} )^{20}$ does not contain any term involving $x^9$

1850904_1709515_ans_338286969fc44adda4d927e6d2c25d5b.jpeg

Show that the term containing $x^3$ does not exist in the expansion of $(3 x -\frac {1}{2x})^8$ .

1850902_1709511_ans_ff06c564fef847ebb616d642b82e8f78.jpeg

Find the coefficient of $x^{-15}$ in the expansion of $(3x^2 - \frac {a}{3x^3} )^{10}$ .

1910052_1709450_ans_072a54512ce5463497fc5331707fc29c.jpeg

Find the coefficient of x in the expansion of $( 1 -3x +7x^2)(1-x)^{16}.$

1909751_1709443_ans_a674ea0efd8b4e468b3705267acc4442.jpeg

Write the general term in the expansion of $( x^2 -y)^6$

It is known that the general term

$T_{r+1}$

$\{$ which is the

$(r+1)^{th}$ term

$\}$ in the binomial expansion of

$(a+b)^n$ is given by

$T_{r+2} = \ ^nC_r a^{n+r} b^r$

Thus, the general term in the expansion of

$(x^2 - y^6)$ is

$T_{r+1} = \ ^6C_r (x^2)^{6-r} (-y)^r = (-1)^r\ ^6C_r . x^{12-2r} . y^r$

Find the coefficient of $x^5$ in the expansion of $(x+ 3)^8$ .

1909757_1709446_ans_6b12a1edbf7e497eb24478f41e7cccd1.jpeg

Find the coefficient of $x^6$ in the expansion of $(3x^2 - \frac {1}{3x})^9$ .

1909761_1709447_ans_e31d0e7cc4204eff9ffb4e834392d424.jpeg

Find the two middle terms in the expansion of $( x^2 +a^2)^5$

1846907_1709542_ans_3e874733aa4e43078c0900988032e4d0.jpeg

Find the middle term in the expansion of :
$(\frac {x}{a} - \frac {a}{x})^{10}$

1846683_1709538_ans_71a0ac16b7db45d3b0d9a42548b93f9a.jpeg

Find the middle term in the expansion of :
$( x^2 - \frac {2}{x})^{10}$

1846860_1709540_ans_a2f88baa3c384a349c45f3a74538d8f9.jpeg

Find the 4th term from the end in expansion of $( \frac {4x}{5} - \frac {5}{2x})^9$

1850689_1709529_ans_e07e659c01584340a4109e77beb284e8.jpeg

Find the two middle terms in the expansion of $( 3x - \frac {x^3}{6})^9$

1847313_1709548_ans_9d850421510745069306e149eda5a538.jpeg

Find the two middle terms in the expansion of $( x^4 - \frac {1}{x^3})^{11}$

1847053_1709543_ans_b6efd0f6c67c42489df8e00b1f10c382.jpeg

Find the 5th term from the end in expansion of $( x- \frac {1}{x})^{12}$ .

1850693_1709526_ans_28ed38d313c44a419861880844ae9f4f.jpeg

Find the middle term in the expansion of $( 3+ x)^6$

1850687_1709536_ans_cb3ae804b6214c8cb3870a7797e87b89.jpeg

Find the middle term in the expansion of $(\frac {x+ 3}2 +3y)^8$

1848547_1709537_ans_2a823c18e62644189e91a205b5470b87.jpeg

Find the two middle terms in the expansion of
$(\frac {p}{x} + \frac {x}{p})^9$

1846721_1709546_ans_25853289c093485f9cd0bf2667cd9ab3.jpeg

Find the coefficient of $x^4$ in the expansion of $( 1+ x)^n(1 -x)^n.$ deduce that $C_2 = C_0C_4 - C_1C_3 +C_2C_2 - C_3C_1 +C_4C_0,$ where $C_r$ stands for $^nC_r.$

1848606_1709571_ans_726c91da629d4dc8a2333db36ba279e0.jpeg

Find a and b so that the $n^{th}$ term in the expansion of $\dfrac {a + bx}{(1 - x)^2}$ may be $(3n - 2)x ^{n - 1}.$

1992575_1744488_ans_acf94c9487a149519ab2e24420a2d67d.jpg

The larger of $99^{50}+100^{50}$ is $101^{50}$ is ________.

We have,

$101^{50}=(100+1)^{50}=100^{50}+50\times 100^{49}+\dfrac {50\times 49}{2\times 1} 100^{48}+...(1)$

$99^{50}=(100-1)^{50}=100^{50}-20\times 100^{49}+\dfrac {50\times 49}{2\times 1}100^{48}-...$
Subtracting (2) from (1), we get

$101^{50}-99^{50}=100^{50}+2\dfrac {50\times 49\times 48}{1\times 2\times 3}100^{47}+... > 100^{50}$
Hence,

$101^{50} > 100^{50}+99^{50}$ .

If $(1+ax)^{n}=1+8x+24x^{2}+....$ then $a=$ _ and $n=$ _

$(1+ax)^n=1+8x+24x^2+...$

$\Rightarrow \ 1+nxa+\dfrac {n(n-1)}{2!}a^2x^2+... =1+8x+24x^2+...$
Comparing like powers of

$x$ , we get

$\dfrac {n(n-1)a^2}{2}=24\Rightarrow n(n-1)a^2=48$
Solving (1) and (2),

$n=4, a=2$ .

Find $a , b , c$ so that the coefficient of $x^{n}$ in the expansion of $\dfrac {a + bx + cx^2}{(1 - x)^3}$ may be $n^2 + 1.$

1992577_1744489_ans_b67538b48af049d88e301707007c40cc.jpg

Prove that the sum of the coefficient of the odd powers of x the expansion of $(1+x+x^2+x^3+x^4)^{n-1}$ , when $n$ is a prime number other than $5,$ is divisible by $n.$

1574078_1745517_ans_c51a70a91e6242558a79a28343b50fcb.jpg

Match the statements $a,b,c,d$ in column I with statements $p,q,r,s$ in column II.

(A) Expansion of

$(1+x)^{41}$ is,

$(1+x)^{41}= ^{41}C_0 +\ ^{41}C_1 x +\ ^{41}C_2 x^2 +...+\ ^{41}C_{20} x^{20}+ ^{41}C_{21}x^{21}+\ ^{41}C_{22}x^{22}+ ... +\ ^{41}C_{41}x^{41}$ ...(1)

$\therefore\$ Sum of binomial coefficients of terms containing power of

$x$ more than

$x^20$

$=^{41}C_{21}+\ ^{41}C_{22}+ ... +\ ^{41}C_{41}$

Sustituting

$x=1$ in equation (1), we get,

$2^{41}= ^{41}C_0 +\ ^{41}C_1 +\ ^{41}C_2 +...+\ ^{41}C_{20} + ^{41}C_{21}+\ ^{41}C_{22}+ ... +\ ^{41}C_{41}$

$\Rightarrow 2^{41}=\ ^{41}C_{41}+\ ^{41}C_{40}+\ ^{41}C_{39}+...+\ ^{41}C_{21}+ ^{41}C_{21}+\ ^{41}C_{22}+ ... +\ ^{41}C_{41}\quad[\because \ ^nC_r=^nC_{n-r}]$

$\Rightarrow 2^{41}=2( ^{41}C_{21}+\ ^{41}C_{22}+ ... +\ ^{41}C_{41})$

$\Rightarrow ^{41}C_{21}+\ ^{41}C_{22}+ ... +\ ^{41}C_{41}=2^{40}$

(B)

$(1+\sqrt 2)^{42} = ^{42}C_0+\ ^{42}(C_1\sqrt 2) +\ ^{42}C_2\ (\sqrt 2)^2 +\ ^{43}C_1 (\sqrt 2)^2 +...+\ ^{42}C_{42} (\sqrt 2)^{42}$

$\therefore$ Sum of binomial coefficients of rational terms is

$=\ ^{42}C_0 +\ ^{42}C_2 +\ ^{42}C_4 +....+\ ^{42}C_{42} (\sqrt 2)^{42}$

We know that,

$\ ^nC_0+\ ^nC2+\ ^nC_4+...=\ ^nC_1+\ ^nC_3+\ ^nC_5+... =2^{n-1}$

$\therefore \ ^{42}C_0 +\ ^{42}C_2 +\ ^{42}C_4 +....+\ ^{42}C_{42} (\sqrt 2)^{42} =2^{41}$

$\left(x+\dfrac {1}{x}+ x^2 + \dfrac {1}{x^2}\right)^{21}= \left(\dfrac {x^3 +x+x^4 +1}{x^2}\right)^{21}=a_0x^{-42} +a_1 x^{-41}+a_2 x^{-40} +...+ a_{82} x^{42}{x^{42}}$ ...(3)

Now, putting

$x=1$ in equation (3), we get

$4^{21}= a_0 +a_1 +a_2 +...+ a_{82}$

Putting

$x=-1$ in equation (3), we get

$0=a_{0} -a_{1}+a_{2} -a_{3}+...+a_{82}$

Adding the above two equations, we get

$4^{21}=2(a_{0}+a_{2}+....+a_{82})$

$\Rightarrow \ a_{0}+a_{2}+...+ a_{82}=2^{41}$

(D) Expansion of

$(1+ix)^{42}$ is,

$(1+ix)^{42}= ^{42}C_0+\ ^{42}C_1(ix) +\ ^{42}C_2\ (ix)^2 +\ ^{43}C_1 (ix)^2 +...+\ ^{42}C_{42} (ix )^{42}$

$\therefore$ Sum of binomial coefficients of positive real terms is,

$=\ ^{42}C_0 +\ ^{42}C_4 +\ ^{42}C_8 +...$

We know that

$\ ^nC_0 -\ ^nC_2 +\ ^nC_2 +\ ^nC_4 -\ ^nC_6 +... =2^{n/2} \cos \dfrac {n \pi}{4}$

and

$\ ^nC_0 +\ ^nC_2 +\ ^nC_4 +\ ^nC_6 +..... =2^{n-2}$

$\Rightarrow \ ^nC_2 +\ ^nC_4 +\ ^nC_8 +... =\dfrac 12 \left(2^{n/2} \times \cos \dfrac {n \pi}{4}+2^{n-1}\right)$

Here,

$n=42$ ,

$^{42}C_0 +\ ^{42}C_4 +\ ^{42}C_8 +... =\dfrac {1}{2} \left(2^{21}\times \cos \dfrac {21 \pi}{2}+2^{41}\right)=2^{40}$

Shew that the middle term in the expansion of $(1+x)^{2n}$ is $\dfrac{1.3.5...(2n-1)}{(n)!}2^nx^n$ .

Number of terms

$2n$ which is even

So,

Middle term

$=\left(\dfrac{2n}{2}+1\right)^{th}$ term

$=(n+1)^{th}$ term

Hence, we need to find

$T_{n+1}$

For

$T_{n+1},$

Putting

$n=2n,\,r=n,\,a=1$ and

$b=x$

$T_{n+1}=\,^{2n}C_n(1)^{2n-n}(x)^n$

$=\dfrac{(2n)!}{n!(2n-n)!}.(1)^n.x^n$

$=\dfrac{(2n)!}{n!n!}.x^n$

$=\dfrac{2n(2n-1)(2n-2)....4\times 3\times 2\times 1}{n!n!}x^n$

$=\dfrac{[(2n-1)(2n-3)...\times 5\times 3\times 1][(2n)(2n-2)...\times 4\times 2]}{n!n!}x^n$

$=\dfrac{[1\times 3\times 5\times ...\times (2n-3)(2n-1)][2\times 4\times 6...\times (2n-2)\times 2n]}{n!n!}x^n$

$=\dfrac{[1\times 3\times 5...\times (2n-3)(2n-1)][(2\times 1)\times(2\times 2)\times (2\times 3)\times ...\times 2(n-1)\times 2n]}{n!n!}x^n$

$=\dfrac{[1\times 3\times 5....\times (2n-3)(2n-1)(2\times 2\times 2\times 2...\times 2)[1\times 2\times 3...(n-1)n}{n!n!}x^n$

$=\dfrac{[1\times 3\times 5...\times (2n-3)(2n-1)]2^n[1\times 2\times 3...(n-1)n]}{n!n!}x^n$

$=\dfrac{[1\times 3\times 5...\times (2n-3)(2n-1)]2^n(n!)}{n!n!}x^n$

$=\dfrac{1\times 3\times 5...\times (2n-1)}{n!}2^n.x^n$

Find the coefficient of $x^n$ in the expansion of
$(1 - 2x + 3x^2 - 4x^3 + \dots)^{-n}$ .

1583579_1743871_ans_a785b7e8e1254a8494f623e05d4ef426.jpg

Find n, if the ratio of fifth term from the beginning to the fifth term from the end in the expansion of $\left ( \sqrt[4]{2} + \cfrac{1}{\sqrt[4]{3}} \right )^{n} \text {is} \sqrt{6} : 1$

We have

$T_{r + 1} = \,^{n}C_{r} (\sqrt[4]{2} )^{n - r} \left ( \dfrac{1}{\sqrt[4]{2}} \right )^{r}$
Fifth term from the beginning

$= T_{4 + 1}$

$= \,^{n}C_{4} (\sqrt[4]{2})^{n - 4} \left ( \dfrac{1}{\sqrt[4]{3}} \right )^{4}$
Fifth term from the end

$= T_{n - 5 + 2} = T_{n - 3}$

$= \,^{n}C_{n - 4} (\sqrt[4]{2})^{n - (n - 4)} \left ( \dfrac{1}{\sqrt[4]{3}} \right )^{n - 4}$

$= \,^{n}C_{n - 4} (\sqrt[4]{2} )^{4} \left ( \dfrac{1}{\sqrt[4]{3}} \right )^{n - 4}$
Given, ratio of fifth term from beginning and end

$= \dfrac{\sqrt 6}{1}$

$\Rightarrow \dfrac{\,^{n}C_{4}2^{\dfrac{n - 4}{4}} \left ( \dfrac{1}{\sqrt[4]{3}} \right )^{4}}{\,^{n}C_{4} 2 \left ( \dfrac{1}{\sqrt[4]{3}} \right )^{n - 4}} = \dfrac{\sqrt 6}{1}$

$\Rightarrow 2^{\dfrac{n - 4}{4} - 1} \left ( \dfrac{1}{3} \right )3^{\dfrac{n - 4}{4}} = \sqrt{6}$

$\Rightarrow 2^{\dfrac{n - 8}{4}} . 3^{\dfrac{n - 8}{4}} = 6^{\dfrac{1}{2}}$

$\therefore \dfrac{n - 8}{4} = \dfrac{1}{2} \Rightarrow n - 8 = 2$

$\therefore n = 10.$

Find the $4^{th}$ term in the expansion of $(x - 2y)^{12}.$

$We\space have\space$

$T_{r + 1} = ^{n}C_{r} \,a^{n - r}b^{r}$

$Here$

$r + 1 = 4 \Rightarrow r = 3$

$n = 12, a = x, y = -2y$

$\therefore T_{4} = ^{12}C_{3}x^{12 - 3} (-2y)^{3}$

$= \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1} . x^{9} (-8y^{3}) = -1760 x^{9}y^{3}$

$9th$ term of $\Bigg(\dfrac{x}{y} - \dfrac{3y}{x^2}\Bigg)^{12}$

$9th$ term of

$\Bigg(\dfrac{x}{y} - \dfrac{3y}{x^2}\Bigg)$

$\because\,$ We know that

$T_{r+1}=\,^nC_r\,a^{n-r}\,b^r$

$Here,\,a=x/y,\,b=\dfrac{-3y}{x^2},\,n=12,r=8$

Now

$9th$ term

$=T_9=T_{8+1}$

$\Rightarrow T_9=\,^{12}C_8\,\bigg(\dfrac{x}{y}\bigg)^{12-8}\bigg(-\dfrac{3y}{x^2}\bigg)^{8}$

$\Rightarrow T_9=495 \times \dfrac{x^4}{y^4} \times \bigg(\dfrac{3^8y^8}{x^{16}}\bigg)$

$\Rightarrow T_9=495 \times 6561 \times \dfrac{y^{8-4}}{x^{16-4}}$

$\Rightarrow T_9=\dfrac{3247695y^4}{x^{12}}$

Hence,

$9th$ term

$=\dfrac{3247695y^4}{x^{12}}$

$5th$ term of $(a + 2x^3)^{17}$

$5th$ term of

$(a + 2x^3)^{17}$
We know that

$(r + 1 )^{th}$ term in the expansion of

$(a + b)^n$

$T_{r+1} =\,^nC_r a^{n-r} b^r$
Here

$a = ‘a’, b = 2x^3, n = 17$

$r + 1 = 5 \Rightarrow r = 4$
Hence,

$5th$ term

$= T_5$

$\Rightarrow T_5 = T_4+1$

$\Rightarrow T_5 =\,^{17}C_4\,a^{17-4}\,(2x^3)^4$

$\Rightarrow T_5 = 2380 × a^{13} × 2^4 × x^{12}$

$\Rightarrow T_5 = 2380 \times 16 \times a^{13} x^{12}$

$\Rightarrow T_5 = 38080 a^{13} x^{12}$
Hence,

$5th$ term

$= 38080 a^{13} x^{12}$ or

$\,^{17}C_4 a^{13} \times 16x^{12}$

Find $a$ , if the $17^{th}$ and $18^{th}$ terms of the expansion $(2 + a)^{50}$ are equal.

$We\space have$

$T_{r + 1} = \,^{50}C_{r} 2^{50 - r}a^{r}$

$\therefore T_{17} = \,^{50}C_{16}2^{34}a^{16}$

$T_{17} = \,^{50}C_{16}2^{34}a^{16}$

$T_{18} = \,^{50}C_{17}2^{33}a^{17}$

Given

$T_{17} = T_{18}$

$\therefore ^{50}C_{16}2^{34}a^{16} = ^{50}C_{17}2^{33}a^{17}$

$\Rightarrow \dfrac{^{50}C_{16} . 2^{34}}{^{50}C_{17} . 2^{33}} = \dfrac{a^{17}}{a^{16}}$

$\Rightarrow a = \dfrac{50!}{16! \,34!} \times \dfrac{17! \,33!}{50!} \times 2$

$a= \dfrac{17 \,16! \,33!}{16! . 34 \,33!} \times 2 = 1$

Find the middle terms in the expansions of
(i) $\left (3 - \cfrac{x^{3}}{6} \right )^{7}$
(ii) $\left (\cfrac{x}{3} + 9 y \right)^{10}$

Here

$n = 7$ is odd

$\therefore$ There are two middle terms

$T_{\left ( \dfrac{7 + 1}{2} \right )}$ and

$T_{\left ( \dfrac{7 + 3}{2} \right )}$ i.e.,

$T_{4}$ and

$T_{5}.$
We have

$T_{r + 1} = \,^{7}C_{r}3^{7 - r} \left ( -\dfrac{x^3}{6} \right )^{r}$

$\therefore T_{4} = ^{7}C_{3} . 3^{4} \left ( -\dfrac{x^3}{6} \right )^{3} = - \,^{7}C_{3} . \dfrac{3^4 . x^9}{6^3}$

$= -\dfrac{7(6)5}{3(2)1} . \dfrac{81}{216} . x^9$

$T_{5} = \,^{7}C_{4} . 3^{3} \left ( -\dfrac{x^3}{6} \right )^{4}$

$= \dfrac{7(6) 5(4)}{4(3)2(1)} . \dfrac{27}{36(36)} . x^{12}$

$= \dfrac{35}{48}x^{12}$

Expand the following Binomials upto fourth term: $(1+x^2)^{-2}$

$(1+x^2)^{-2}=1+(-2)x^2+\dfrac{(-2)(-2-1)(x^2)^2}{2!}+\dfrac{(-2)(-2-1)(-2-2)(x^2)^3}{3!}$

$\Rightarrow (1+x^2)^{-2}=1-2x^2+\dfrac{6x^4}{2!}+\dfrac{(-2)(-3)(-4)x^6}{3!}$

$\Rightarrow (1+x^2)^{-2}=1-2x^2+\dfrac{6x^4}{2!}-\dfrac{24x^6}{3!}$

$\Rightarrow (1+x^2)^{-2}=1-2x^2+\dfrac{6x^4}{2 \times 1}-\dfrac{24x^6}{3 \times 2 \times 1}$

$\Rightarrow (1+x^2)^{-2}=1-2x^2+3x^4-4x^6$

Find the $13^{th}$ term in the expansion of
$\left ( 9x - \dfrac{1}{3\sqrt{x}} \right )^{18}, x \neq 0$

$We$

$have$

$T_{r + 1} = ^{n}C_{r}a^{n - r}b^{r}$

$Where$ ,

$r + 1 = 13 \Rightarrow r = 12$

$n = 18, a = 9x, b = -\dfrac{1}{3\sqrt x}$

$\therefore T_{13} = ^{18}C_{12} (9x)^{18 - 12} \left ( \dfrac{-1}{3\sqrt x} \right )^{12}$

$= ^{18}C_{12}9^{6} . x^{6} \dfrac{1}{9^6 . x^6}$

$= ^{18}C_{12}$

Find the general term of the following expansions:
$(1 x)^{-p/q}$

$T_{r+1}=\cfrac{-\dfrac{p}{q}\bigg(-\dfrac{p}{q}-1\bigg)\bigg(-\dfrac{p}{q}-2\bigg)\bigg(-\dfrac{p}{q}-3\bigg).......\bigg(-\dfrac{p}{q}-r+1\bigg)}{r!}(-x)^r$

$=\cfrac{-\dfrac{p}{q}\bigg(\dfrac{-p-q}{q}\bigg)\bigg(\dfrac{-p-2q}{q}-2\bigg)\bigg(\dfrac{-p-3q}{q}-3\bigg).......\bigg(\dfrac{-p-q}{q}\bigg)(-1)^r\,x^r}{r!}$

$=\cfrac{(-1)^rp(p+q)(p+2q)....\{p+(r-1)q\}\, (-1)^r x^r }{q^r \times r!}$

$=\cfrac{p(p+q)(p+2q)....\{p+(r-1)q\}}{r!}\bigg(\dfrac{x}{q}\bigg)^r$

Find the coefficient of $x^6$ in the expansion of $(a + 2bx^2)^{-3}$ .

$(r + 1)th$ term in the expansion of

$(a + 2bx^2)^{-3}$

$T_{r+1}=\dfrac{a^{-3}(-3)(-3-1)(-3-2)....(-3-r+1)}{r!}\bigg(\dfrac{2bx^2}{a}\bigg)^r$

$\Rightarrow T_{r+1}=\dfrac{a^{-3}(-3)(-4)(-5)....(-2-r)}{r!}\bigg(\dfrac{2^rb^rx^{2r}}{a^r}\bigg)$

$\Rightarrow T_{r+1}=\dfrac{(-1)^{r}\,3.4.5....(2+r)\,2^r\,b^r\,x^{2r}}{r! \times a^{r+3}}$

For the coefficient of $x^6$ in the this term $2r = 6 \Rightarrow r = 3$

Hence, from equation (i), coefficient of $x^6$ in the expansion of $(a + 2bx^2)^{-3}$

$= \dfrac{(-1)^3 \times 3 \times 4 \times 5 \times 2^3 \times b^3}{3! \times A^{3+3}}$

$= \dfrac{-3 \times 4 \times 5 \times 8 \times b^3}{3 \times 2 \times a^6}$

$= \dfrac{-80b^3}{a^6}$

$= -80b^{3}a^{-6}$

Find the middle term in the following expansions.
$\bigg(\dfrac{x}{2}+2y\bigg)^6$

$\bigg(\dfrac{x}{2}+2y\bigg)^6$

The middle term in the expansion of

$\bigg(\dfrac{x}{2}+2y\bigg)^6$ . In the expansion of

$\bigg(\dfrac{x}{2}+2y\bigg)^6$ , the middle term will be

$T_{\bigg [\frac{n}{2} + 1\bigg ]th}$ term where

$n = 6$ which is an even number.

The middle term $T_{\bigg [\frac{n}{2} + 1\bigg ]} = T_4$

$T_4 = T_3+1$

$=\,^6C_3\,\bigg[\dfrac{x}{2}\bigg]^{6-3} (2y)^3$

$= 20 \times \bigg[\dfrac{x^3}{2^3}\bigg] \times 2^3 \times y^3$

$= 20 x^3 y^3$

Find the required terms in the following expansions:
Eighth term of $(1+2x)^{-1/2}$

Eighth term of

$(1+2x)^{-1/2}$

$=\dfrac{\bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{-1}{2}-1\bigg)\bigg(\dfrac{-1}{2}-2\bigg)\bigg(\dfrac{-1}{2}-3\bigg)\bigg(\dfrac{-1}{2}-4\bigg)\bigg(\dfrac{-1}{2}-5\bigg)\bigg(\dfrac{-1}{2}-6\bigg)(2x)^7}{7!}$

$=\dfrac{\bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{-3}{2}\bigg)\bigg(\dfrac{-5}{2}\bigg)\bigg(\dfrac{-7}{2}\bigg)\bigg(\dfrac{-9}{2}\bigg)\bigg(\dfrac{-11}{2}\bigg)\bigg(\dfrac{-13}{2}\bigg)\times 2^7 \times x^7}{7!}$

$=\dfrac{-13 \times 9 \times 7 \times 5 \times 3 \times 2^7}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 2^7}x^7$

$=\dfrac{-13 \times 9 \times 7 \times 5 \times 3}{7 \times 6 \times 5 \times 4 \times 3 \times 2}x^7=\dfrac{-429}{16}x^7$

Expand the following Binomials upto fourth term $\bigg(1-\dfrac{x}{2}\bigg)^{1/2}$ .

$\bigg(1-\dfrac{x}{2}\bigg)^{1/2}=\bigg\{1+\bigg(-\dfrac{x}{2}\bigg)\bigg\}^{1/2}$ .

$=1+\dfrac{1}{2}\bigg(\dfrac{-x}{2}\bigg)+\dfrac{\dfrac{1}{2}\bigg(\dfrac{1}{2}-1\bigg)\bigg(\dfrac{-x}{2}\bigg)^2}{2!}+\dfrac{\dfrac{1}{2}\bigg(\dfrac{1}{2}-1\bigg)\bigg(\dfrac{1}{2}-2\bigg)\bigg(\dfrac{-x}{2}\bigg)^3}{3!}$

$=1-\dfrac{x}{4}+\bigg(-\dfrac{1}{4}\bigg) \times \dfrac{x^2}{4} \times \dfrac{1}{2}+\dfrac{1}{2} \times \bigg(-\dfrac{1}{2}\bigg)\bigg(-\dfrac{3}{2}\bigg)\bigg(-\dfrac{x^3}{8}\bigg)\times\dfrac{1}{3\times 2}$

$=1-\dfrac{x}{4}-\dfrac{x^2}{32}-\dfrac{x^3}{128}$

Expand the following Binomials upto fourth term: $(3-2x^2)^{-2/3}$

$(3-2x^2)^{-2/3}=\{3+(-2x^2)\}^{-2/3}$

$(3-2x^2)^{-2/3}=3^{-2/3}\bigg\{1+\bigg(\dfrac{-2x^2}{3}\bigg)\bigg\}^{-2/3}$

$=3^{-2/3}\bigg\{1+\bigg(\dfrac{-2}{3}\bigg)\bigg(\dfrac{-2x^2}{3}\bigg)+\bigg(\dfrac{-2}{3}\bigg)\bigg(\dfrac{-2}{3}-1\bigg)\bigg(\dfrac{-2x^2}{3}\bigg)^2 \times \dfrac{1}{2!}+\bigg(\dfrac{-2}{3}\bigg)\bigg(\dfrac{-2}{3}-1\bigg)\bigg(\dfrac{-2}{3}-2\bigg)\bigg(\dfrac{-2x^2}{3}\bigg)^3 \times \dfrac{1}{3!}\bigg\}$

$=3^{-2/3}\bigg\{1+\dfrac{4x^2}{9}+\dfrac{20x^4}{81}+\bigg(\dfrac{-2}{3}\bigg)\bigg(\dfrac{-5}{3}\bigg)\bigg(\dfrac{-8}{3}\bigg)\bigg(\dfrac{-8}{27}\bigg)x^6 \times \dfrac{1}{6}\bigg\}$

$=\dfrac{1}{3^{2/3}}\bigg\{1+\dfrac{4x^2}{9}+\dfrac{20x^4}{81}+\dfrac{320}{2187}x^6\bigg\}$

Expand the following Binomials upto fourth term: $\dfrac{1}{\sqrt{5+4x}}$ .

$\dfrac{1}{\sqrt{5+4x}}=\dfrac{1}{(5+4x)^{1/2}}=\dfrac{1}{5^{1/2}\bigg(1+\dfrac{4}{5}x\bigg)^{1/2}}$

$=\dfrac{1}{5^{1/2}}\bigg(1+\dfrac{4}{5}x\bigg)^{-1/2}$

$=\dfrac{1}{5^{1/2}}\bigg[1+\bigg(\dfrac{-1}{2}\bigg)\dfrac{4}{5}x+\bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{-1}{2}-1\bigg)\bigg(\dfrac{4}{5}x\bigg)^2+\bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{-1}{2}-1\bigg)\bigg(\dfrac{-1}{2}-2\bigg)\bigg(\dfrac{4}{5}x\bigg)^3.\dfrac{1}{3!}\bigg]$

$=\dfrac{1}{5^{1/2}}\bigg[1-\dfrac{2x}{5}+\bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{-3}{2}\bigg)\bigg(\dfrac{16}{25}x^2\bigg)^2 \times \dfrac{1}{2}+\bigg(\dfrac{-1}{2}\bigg)\bigg(\dfrac{-3}{2}\bigg)\bigg(\dfrac{-3}{2}\bigg)\bigg(\dfrac{64}{125}x^3\bigg).\dfrac{1}{6}\bigg]$

$=\dfrac{1}{5^{1/2}}\bigg[1-\dfrac{2x}{5}+\dfrac{6}{25}x^2- \dfrac{4}{25}x^3\bigg]$

$=\dfrac{1}{\sqrt{5}}\bigg[1-\dfrac{2x}{5}+\dfrac{6}{25}x^2- \dfrac{4}{25}x^3\bigg]$

Find the middle term in the following expansions.
$\bigg(x+\dfrac{1}{x}\bigg)^{2n}$

Here $n = 2n$ , which is an even number.

The middle term is $\Bigg \{\frac{2n}{2} + 1\Bigg \}th$ term and $\{(n + 1)\}th$ term.

Now, $T_{r+1} =\,^{2n}C_r\,(x)^{2n-r} \bigg(\dfrac{1}{x}\bigg)^r$

Put $r = n$ ,

$T_{n+1} =\,^{2n}C_n\,(x)^{2n-n} \bigg(\dfrac{1}{x^n}\bigg)$

$= \dfrac{(2n)!}{n! \times n!} \times x^n \times \bigg(\dfrac{1}{x^n}\bigg)$

$= \dfrac{(2n)!}{(n!)^2}$

The middle term is $\dfrac{(2n)!}{(n!)^2}$ .

Seventh term of $(1+x)^{5/2}$

Seventh term of

$(1+x)^{5/2}$

$=\dfrac{\dfrac{5}{2}\bigg(\dfrac{5}{2}-1\bigg)\bigg(\dfrac{5}{2}-2\bigg)\bigg(\dfrac{5}{2}-3\bigg)\bigg(\dfrac{5}{2}-4\bigg)\bigg(\dfrac{5}{2}-5\bigg)x^6}{6!}$

$=\dfrac{\dfrac{5}{2} \times \dfrac{3}{2} \times \dfrac{1}{2} \times \bigg(\dfrac{-1}{2}\bigg) \times \bigg(\dfrac{-3}{2}\bigg) \times \bigg(\dfrac{-5}{2}\bigg)\times x^6}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$=\dfrac{-5 \times 3 \times 3 \times 5}{64 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}x^6=\dfrac{-5 \times 3 \times 15}{64 \times 3 \times 2 \times 15 \times 8}x^6$

$=-\dfrac{13 \times 11 \times 9}{6 \times 4 \times 2}=\dfrac{-5}{1024}x^6$

Find the required terms in the following expansions: Fourth term of $(1 3x)^{-1/3}$ .

Fourth term of

$(1 - 3x)^{-1/3}$
= Fourth term of

$(1 +(-3x))^{-1/3}$

$=\dfrac{\bigg(-\dfrac{1}{3}\bigg)\bigg(-\dfrac{1}{3}-1\bigg)\bigg(-\dfrac{1}{3}-2\bigg)(-3x)^3}{3 \times 2}$

$=\dfrac{\bigg(-\dfrac{1}{3}\bigg)\bigg(-\dfrac{4}{3}\bigg)\bigg(-\dfrac{7}{3}\bigg)(-27)x^3}{3 \times 2}$

$=\dfrac{4 \times 7 \times 27x^3}{27 \times 3 \times 2}=\dfrac{14}{3}x^3$

The ratio of fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt[4]{2} + \dfrac{1}{\sqrt[4]{3}}\right)^{n}$ is $\sqrt{6} : 1$ . If $n = \dfrac{20}{\lambda}$ , find the value of $\lambda$ .

${T}_{4+1}$ from begining is

$^{n}{C}_{4}$ .

$\cfrac { { 2 }^{ \frac { n-4 }{ 4 } } }{ { 3 }^{ \cfrac { 4 }{ 4 } } }$

And

${T"}_{4+1}$ from end

$^{n}{C}_{n-4}$ .

$\cfrac { { 2 }^{ \cfrac { 4 }{ 4 } } }{ { 3 }^{ \cfrac { n-4 }{ 4 } } }$

$\Rightarrow$

$\cfrac { { T }_{ 5 } }{ { T" }_{ 5 } }$ =

$\sqrt { 6 }$

$\Rightarrow$

$\cfrac { { 2 }^{ \frac { n-4 }{ 4 } }.{ 3 }^{ \cfrac { n-4 }{ 4 } } }{ { 2 }.{ 3 } }$ =

${ 6 }^{ \cfrac { 1 }{ 2 } }$

$\Rightarrow$

$\cfrac { { 6 }^{ \cfrac { n-4 }{ 4 } } }{ 6 }$ =

${ 6 }^{ \cfrac { 1 }{ 2 } }$

$\Rightarrow$

${ 6 }^{ \cfrac { n-4 }{ 4 } }$ =

${6}^{\cfrac{3}{2}}$

hence

$\Rightarrow$

$\dfrac{n-4}{4}$ =

$\dfrac { 3 }{ 2 }$

$\Rightarrow$ n-4=6

$\Rightarrow$ n=10

$\Rightarrow$

$\lambda$ =

$\dfrac{n}{2}$ so

$\lambda$ =2

Find the coefficient of ${ x }^{ 50 }$ in the polynomials after parenthesis have been removed and like terms have been collected in the expansion
${ \left( 1+x \right) }^{ 1000 }+x{ \left( 1+x \right) }^{ 999 }+{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+....+{ x }^{ 1000 }$

Given expansion is

${\left( 1+x \right) }^{ 1000 }+2x{ \left( 1+x \right) }^{ 999 }+3{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+3{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+....+1001{ x }^{ 1000 }$

Let

$S$ denotes the sum of the polynomial.

$S= { \left( 1+x \right) }^{ 1000 }+2x{ \left( 1+x \right) }^{ 999 }+3{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+3{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+....+1001{ x }^{ 1000 }$ .....(1)

Multiplying by

$\displaystyle \frac{x}{1+x}$ in eqn (1), we get

$\displaystyle \frac { x }{ (1+x) } S= x{ \left( 1+x \right) }^{ 999 }+2{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+3{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+...+1000{ x }^{ 1000 }+\frac { 1001 }{ (1+x) } { x }^{ 1001 }$ ....(2)

Subtracting eqn(2) from eqn(1), we get

$\displaystyle \left( 1-\frac { x }{ 1+x } \right) S=(1+x)^{ 1000 }+x{ \left( 1+x \right) }^{ 999 }+{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+...+{ x }^{ 1000 }-\frac { 1001 }{ (1+x) } { x }^{ 1001 }$

$\Rightarrow \displaystyle \left( \frac { 1 }{ 1+x } \right) S=(1+x)^{ 1000 }+x{ \left( 1+x \right) }^{ 999 }+{ x }^{ 2 }{ \left( 1+x \right) }^{ 998 }+{ x }^{ 3 }{ \left( 1+x \right) }^{ 997 }+....+{ x }^{ 1000 }-\frac { 1001 }{ (1+x) } { x }^{ 1001 }$

$\Rightarrow \displaystyle S=[(1+x)^{ 1001 }+x{ \left( 1+x \right) }^{ 1000 }+{ x }^{ 2 }{ \left( 1+x \right) }^{ 999 }+{ x }^{ 3 }{ \left( 1+x \right) }^{ 998 }+....+{ x }^{ 1000 }(1+x)]-1001{ x }^{ 1001 }$

The sequence in square bracket forms a G.P.

Here, first term

$a=(1+x)^{ 1001 }$ and common ratio

$r=\cfrac{x}{1+x}$ .

So, using the formula for sum of n terms of G.P.

$S_n=\cfrac{a(1-r^n)}{1-r}$

$\Rightarrow \displaystyle S=\frac { (1+x)^{ 1001 }[1-\left( \cfrac { x }{ 1+x } \right) ^{ 1001 }] }{ 1-\cfrac { x }{ 1+x } } -1001{ x }^{ 1001 }$

$\Rightarrow \displaystyle S=\frac { (1+x)^{ 1001 }[1-\left( \cfrac { x }{ 1+x } \right) ^{ 1001 }] }{ \cfrac { 1 }{ 1+x } } -1001{ x }^{ 1001 }$

$\Rightarrow \displaystyle S=(1+x)^{ 1002 }[1-\left( \frac { x }{ 1+x } \right) ^{ 1001 }]-1001{ x }^{ 1001 }$

$\Rightarrow \displaystyle S=(1+x)^{ 1002 }-(1+x)x^{ 1001 }-1001{ x }^{ 1001 }$

$\Rightarrow S=(1+x)^{ 1002 }-1002x^{ 1001 }-x^{ 1002 }$

Since, the last two terms in the above sum cannot have

$x^{50}$ and only the first term can have

$x^{50}$

Now, the coefficient of

$x^r$ in the expansion of

$(1+x)^n$ is

$^nC_r$
So, the coefficient of

$x^{50}$ in

$(1+x)^{1002}$ is

$^{1002}C_{50}$

If ${ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+....+{ C }_{ n }{ x }^{ n }$ , then find the sum of the series
$\cfrac { { C }_{ 0 } }{ 2 } -\cfrac { { C }_{ 1 } }{ 6 } +\cfrac { { C }_{ 2 } }{ 10 } -\cfrac { { C }_{ 3 } }{ 14 } +........+\cfrac { { \left( -1 \right) }^{ n }{ C }_{ n } }{ 4n+2 }$

Given,

$(1+x)^n=C_0+C_1x+C_2x^2+......+C_nx^n$

Consider

$\int (1-x^2)^n dx=\int (C_0+C_1(-x^2)+C_2(-x^2)^2+.....+(-1)^nC_nx^{2n})$

$\Rightarrow \int (1-x^2)^n dx = (C_0-\dfrac{C_1x^3}{3}+.......+\dfrac{(-1)^nC_nx^{2n+1}}{2n+1})$

Substitute

$x=1$ in above equation:

Required sum of series is

$\dfrac{ \int (1-x^2)^n dx}{2}$ at

$x=1$ .

For $n\ge 2$ , let ${ C }_{ r }=\begin{pmatrix} n \\ r \end{pmatrix}$ and ${ a }_{ n }=\sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { C }_{ r } } }$

A) Using

${ C }_{ r }={ C }_{ n-r }$

$\sum _{ r=0 }^{ n }{ \cfrac { r }{ { C }_{ r } } } =\sum _{ r=0 }^{ n }{ \cfrac { r }{ { C }_{ n-r } } } =\sum _{ r=0 }^{ n }{ \cfrac { n-r }{ { C }_{ r } } } =\sum _{ r=0 }^{ n }{ \cfrac { n }{ { C }_{ r } } } -\sum _{ r=0 }^{ n }{ \cfrac { r }{ { C }_{ r } } } \\ \Rightarrow 2\sum _{ r=0 }^{ n }{ \cfrac { r }{ { C }_{ n-r } } } =\sum _{ r=0 }^{ n }{ \cfrac { n }{ { C }_{ r } } } =n{ a }_{ n }\Rightarrow \quad \sum _{ r=0 }^{ n }{ \cfrac { r }{ { C }_{ n-r } } } =\sum _{ r=0 }^{ n }{ \cfrac { r }{ { C }_{ r } } } =\cfrac { 1 }{ 2 } n{ a }_{ n }$

B)

$\sum _{ r=0 }^{ n }{ \cfrac { n-r }{ { C }_{ n-r } } } =\sum _{ r=0 }^{ n }{ \cfrac { n }{ { C }_{ r } } } -\sum _{ r=0 }^{ n }{ \cfrac { r }{ { C }_{ r } } } =n{ a }_{ n }-\cfrac { 1 }{ 2 } n{ a }_{ n }=\cfrac { 1 }{ 2 } n{ a }_{ n }$

C) Using

$\quad { rC }_{ r }=n{ C }_{ r-1 }$

$\sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { rC }_{ r } } } =\cfrac { 1 }{ n } \sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { C }_{ r-1 } } } =\cfrac { 1 }{ n } \sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { C }_{ r } } } =\cfrac { 1 }{ n } { a }_{ n-1 }$

D)

$\sum _{ r=0 }^{ n-1 }{ \cfrac { 1 }{ { (n-r)C }_{ r } } } =\sum _{ r=0 }^{ n-1 }{ \cfrac { 1 }{ { n{ C }_{ r }-rC }_{ r } } } =\sum _{ r=0 }^{ n-1 }{ \cfrac { 1 }{ n{ C }_{ r }-n{ C }_{ r-1 } } } =\cfrac { 1 }{ n } { a }_{ n-1 }$

Match the entries in column I with entries in column II

Consider D,
For the sum of the coefficients, we substitute

$x=1$ , we get

$(1+2-2)^{2009}$

$=1$ .
Hence sum of the coefficients is

$1$ .

Consider C
Total number of rational terms

$=\dfrac{10}{LCM(2,5)}+1$

$=2$ .
Hence total number of rational terms are

$2$ which are

$T_{1}$ and

$T_{11}$

Consider B

$(1+x)^{m}(1+x)^{n}$

$=(1+x)^{m+n}$

$=1+\:^{m+n}C_{1}x+\:^{m+n}C_{2}x^{2}+...\:^{m+n}C_{m+n}x^{m+n}$

Coefficient of

$x$

$=\:^{m+n}C_{1}$

$m+n=3$ ...(i)
Hence

$m+n=3$ and
Coefficient of

$x^{2}$

$=\:^{m+n}C_{2}=-6$
Or

$\dfrac{(m+n)(m+n-1)}{2}=-6$

$(m+n)(m+n-1)=-12$

$(m+n)^{2}-(m+n)+12=0$ ...(ii)
As obtained from equation i

$m+n=3$ .
This does not satisfy the equation no, ii.
Hence no solution for

$m$ and

$n$ .

Consider A,
The degree of the polynomial is

$7$ .

If the 2nd,3rd and 4th terms in the expansion of ${(a+x)}^{n}$ are respectively $240,720,1080$ , find $a,x,n$ .

Given that

${ ^{ n }{ C } }_{ 1 }{ x }^{ n-1 }a=240$ ,

${ ^{ n }{ C } }_{ 2 }{ x }^{ n-2 }{ a }^{ 2 }=720$ ,

${ ^{ n }{ C } }_{ 3 }{ x }^{ n-3 }{ a }^{ 3 }=1080$

By dividing second equation by first equation , we get

$(n-1)\dfrac{a}{x}=6$

By dividing third equation by second equation , we get

$(n-2)\dfrac{a}{x}=\dfrac{9}{2}$

By dividing above two ones , we get

$\dfrac{n-2}{n-1}=\dfrac{9}{12}=\dfrac{3}{4}$

Therefore

$4n-8=3n-3$ , which implies

$n=5$ and

$\dfrac{a}{x} = \dfrac{6}{4} = \dfrac{3}{2}$

from first equation , we get

${ ^{ 5 }{ C } }_{ 1 }{ x }^{ 4 }{ a }=240$

$\Rightarrow {x}^{4}a=48$

we know that

$\dfrac{a}{x}=\dfrac{3}{2}$ , which gives

$a=\dfrac{3x}{2}$

By substituting value of

$a$ in

${x}^{4}a=48$ , we get

${x}^{5}=32$ , which gives

$x=2$ and

$a=3$

Find the middle term in the expansion of ${\left(\dfrac{a}{x}+bx\right)}^{12}$

1339870_427247_ans_40482654accf4fbbbbc1d59fdc6e7f66.PNG

If ${c}_{0},{c}_{1},{c}_{2},.......{c}_{n}$ denote the coefficients in the expansion of ${(1+x)}^{n}$ , prove that
$\cfrac { { c }_{ 1 } }{ { c }_{ 0 } } +\cfrac { 2{ c }_{ 2 } }{ { c }_{ 1 } } +\cfrac { 3{ c }_{ 3 } }{ { c }_{ 2 } } +...\cfrac { n{ c }_{ n } }{ { c }_{ n-1 } } =\cfrac { n(n+1) }{ 2 }$ .

$S_n = \dfrac{c_1}{c_0} +\dfrac{2c_2}{c_1} +\dfrac{3c_3}{c_2} +\ldots \ldots \dfrac{nc_n}{c_{n-1}}$

$T_r = r\dfrac{c_r}{c_{r-1}} = r\cdot \dfrac{n!}{(n-r)!r!} \cdot \dfrac{(n-r+1)!(r-1)!}{n!} = n+1-r$

$R.H.S. = S_n = \displaystyle\sum_{r=1}^n T_n = \sum_{r=1}^n (n+1-r) = n^2+n - \dfrac{n(n+1)}{2} = \dfrac{n(n+1)}{2} = L.H.S.$

Find the two middle terms of ${ \left( 3a-\cfrac { { a }^{ 3 } }{ 6 } \right) }^{ 9 }$ .

There will be

$10$ terms.
So, we have two middle terms

${ T }_{ 5 }$ and

${ T }_{ 6 }$

General term

$T_{r+1}$ in the expansion of

$(x-y)^{n}$ is given by

$T_{r+1}=(-1)^{r} \ {}^{n}C_{r} x^{n-r} y^{r}$

$\therefore { T }_{ 5 }={}^{ 9 }{ C }_{ 4 }{ \left( 3a \right) }^{ 5 }{ \left( -\cfrac { { a }^{ 3 } }{ 6 } \right) }^{ 4 }$

$=\cfrac { 189 }{ 8 } { a }^{ 17 }$

${ T }_{ 6 }={}^{ 9 }{ C }_{ 5 }{ \left( 3a \right) }^{ 4 }{ \left( -\cfrac { { a }^{ 3 } }{ 6 } \right) }^{ 5 }$

$=-\cfrac { 21 }{ 16 } { a }^{ 19 }$

If $p$ is a prime number, show that the coefficients of the terms of $(1 + x)^{p - 1}$ are alternately greater and less by unity than some multiple of $p$ .

Let the co efficients be denoted by

${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },.........{ C }_{ r },,..$

then

${ C }_{ r }=\cfrac { \left( p-1 \right) \left( p-2 \right) \left( p-3 \right) ...\left( p-r \right) }{ r! }$

$=\cfrac { M\left( p \right) +\left( -1 \right) r! }{ r! } =\cfrac { M(p) }{ r! } +{ \left( -1 \right) }^{ r }$

Now,

${ C }_{ r }$ is a positive integer;hence

$\cfrac { M(p) }{ r! } +{ \left( 1 \right) }^{ r }$ must be a positive integer, and since p is a prime number it must be a multiple of p; therefore

${ C }_{ r }=M(p)+{ \left( -1 \right) }^{ r }$

If $c_0 , c_1 , c_2 , ...... c_n$ are the coefficients in the expansion $(1+x)^n ,$ where $n$ is a positive integer, shew that

$c_1 - \dfrac {c_2}{2} + \dfrac{c_3}{3} - ...... + \dfrac { (-1)^{n-1} c_n }{n} = 1 + \dfrac {1}{2} + \dfrac {1}{3} + .... \dfrac {1}{n}$

Let

$S_n = c_1 - \dfrac{c_2}{2} + \dfrac{c_3}{3} - \ldots \ldots + \dfrac{(-1)^{n-1}c_n}{n}$

We know the coefficients of the expansion of

$(1+x)^n$

$\therefore S_n = n - \dfrac{n(n-1)}{2\cdot 2!} + \dfrac{n(n-1)(n-2)}{3\cdot 3!} - \ldots \ldots \text{to n terms}$

Similarly,

$S_{n+1} = (n+1) - \dfrac{(n+1)n}{2\cdot 2!} - \dfrac{(n+1)n(n-1)}{3\cdot 3!} + \ldots \ldots \text{to n+1 terms}$

$S_{n+1} - S_n = 1 - \dfrac{n}{2!} + \dfrac{n(n-1)}{3!} - \ldots \ldots \text{to n+1 terms}$

Also we know that ,

$(1+x)^{n+1} = 1 + (n+1)x + \dfrac{(n+1)n}{2!}x^2 + \dfrac{(n+1)n(n-1)}{3!}x^3 + \ldots \ldots \text{to n+1 terms}$

For

$x=-1, 0 = 1 - (n+1) + \dfrac{(n+1)n}{2!} - \dfrac{(n+1)n(n-1)}{3!} + \ldots \ldots \text{to n+1 terms}$

$\therefore \,\, 1 - \dfrac{n}{2!} + \dfrac{n(n-1)}{3!} - \ldots \ldots \text{to n+1 terms} = \dfrac{1}{n+1}$

From above we can say that

$S_{n+1}- S_n = \dfrac{1}{n+1}$

$\implies S_2 - S_1 = \dfrac{1}{2}$ , but since

$S_1 = 1$ , thus

$S_2 = 1+\dfrac{1}{2}$

Similarly,

$S_3 = S_2 + \dfrac{1}{3} = 1+\dfrac{1}{2}+\dfrac{1}{3}$

Similarly,

$S_n = 1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ldots\ldots +\dfrac{1}{n}$

Hence Proved

If ${c}_{0},{c}_{1},{c}_{2},.......{c}_{n}$ denote the coefficients in the expansion of ${(1+x)}^{n}$ , prove that
${ c }_{ 0 }+\cfrac { { c }_{ 1 } }{ 2 } +\cfrac { { c }_{ 2 } }{ 3 } +......+\cfrac { { c }_{ n } }{ n+1 } =\cfrac { { 2 }^{ n+1 }-1 }{ n+1 }$ .

$(1+x)^{n+1} = 1+(n+1)x + \dfrac{(n+1)n}{1\cdot 2} x^2 + \dfrac{(n+1)n(n-1)}{1\cdot 2\cdot 3} x^3 +\ldots \ldots + (n+1)x^n + x^{n+1}$

$\implies \dfrac{(1+x)^{n+1}-1}{n+1} = x + \dfrac{n}{1\cdot 2} x^2 + \dfrac{n(n-1)}{1\cdot 2\cdot 3} x^3 +\ldots \ldots + \dfrac{x^{n+1}}{n+1}$

$\implies \dfrac{(1+x)^{n+1}-1}{n+1} = x + \dfrac{c_1}{2} x^2 + \dfrac{c_2}{3} x^3 +\ldots \ldots + \dfrac{c_nx^{n+1}}{n+1}$

Substituting

$x = 1$ , we get

$\dfrac{2^{n+1}-1}{n+1} = 1 + \dfrac{c_1}{2} + \dfrac{c_2}{3} +\ldots \ldots + \dfrac{c_n}{n+1}$

Hence Proved

If ${x}^{p}$ occurs in the equation ${ \left( { x }^{ 2 }+\cfrac { 1 }{ x } \right) }^{ 2n }$ ., prove that its coefficient is $\cfrac { |\underline { 2n } }{ |\underline { \cfrac { 1 }{ 3 } (4n-p) } \quad |\underline { \cfrac { 1 }{ 3 } (2n+p) } }$ .

$\Longrightarrow$ General term:-
${ T }_{ r+1 }={}^{ 2n }{ C }_{ r }{ \left( { x }^{ 2 } \right) }^{ 2n-r }{ \left( \cfrac { 1 }{ x } \right) }^{ r }$

$={}^{ 2n }{ C }_{ r }\cdot { x }^{ 4n-3r }$

$4n-3r=p$ for ${ x }^{ p }$

$\therefore r=\cfrac { 4n-p }{ 3 }$

Coefficient $={}^{ 2n }{ C }_{ r }$

$={}^{ 2n }{ C }_{ \cfrac { 4n-p }{ 3 } }$

$=\cfrac { \left( 2n \right) ! }{ \left( \cfrac { 4n-p }{ 3 } \right) !\left( \cfrac { 2n+p }{ 3 } \right) ! }$

If ${c}_{0},{c}_{1},{c}_{2},.......{c}_{n}$ denote the coefficients in the expansion of ${(1+x)}^{n}$ , prove that
$({c}_{0}+{c}_{1})({c}_{1}+{c}_{2}).......({c}_{n-1}+{c}_{n})=\cfrac { { c }_{ 1 }{ c }_{ 2 }...{ c }_{ n }{ (n+1) }^{ n } }{ |\underline { n } }$ .

$({c}_{0}+{c}_{1})({c}_{1}+{c}_{2})\ldots({c}_{n-1}+{c}_{n})=\dfrac { { c }_{ 1 }{ c }_{ 2 }\ldots{ c }_{ n }{ (n+1) }^{ n } }{ n ! }$

This can be rearranged as,

$\left( \dfrac{c_0+c_1}{c_1} \right) \left( \dfrac{c_1+c_2}{c_2} \right) \ldots \ldots \left( \dfrac{c_{n-1}+c_n}{c_n} \right)=\dfrac{(n+1)^n}{n!}$

$\displaystyle \prod_{r=0}^{n-1} \left( \dfrac{c_r+c_{r+1}}{c_{r+1}} \right) = \dfrac{(n+1)^n}{n!}$

Also,

$\left( \dfrac{c_r+c_{r+1}}{c_{r+1}} \right) = \left( \dfrac{c_r}{c_{r+1}} +1 \right) = \dfrac{n!}{(n-r-1)!(r+1)!} \times \dfrac{(n-r)!r!}{n!} + 1 = \dfrac{n+1}{n-r}$

Therefore,

$L.H.S = \displaystyle \prod_{r=0}^{n-1} \left( \dfrac{c_r+c_{r+1}}{c_{r+1}} \right) = \displaystyle \prod_{r=0}^{n-1} \left( \dfrac{n+1}{n-r} \right) = \dfrac{(n+1)^n}{n!} = R.H.S$

Find the middle term(s) in the expansion of $(1+3x+3x^2+x^3)^{2n}$ .

To find :- Middle term in the expansion of

$(1+3x+3x^2+x^3)^{2n}$ .

solution :-

$(1+3x+3x^2+x^3)^{2n}=[(1+x)^3]^{2n}$

$(\because$ we know

$(a+b)^3=a^3+b^3+3a^2b+3ab^2)$

$=(1+x)^{6n}$

Here Power

$=6n=2(3n)=even$

Here Total no. of terms are odd.

So, There are only one middle term

i.e.

$\left(\dfrac{6n}{2}\right)^{th}\ term=(3n)^{th}\ term$

$T_{3n}$ is to be find now,

we know formula for general term.

$T_{r+1}=^nC_{r}x^{n-r}. y^{r}$ in expansion of

$(x+y)^n$ .

$\boxed{T_{3n}=^{6n}C_{(3n-1)}x^{(3n-1)}}\Longleftrightarrow$ final answer

Find the middle terms in the expansion of ${ \left( { x }^{ 2 }+\dfrac { 1 }{ x } \right) }^{ 11 }$

Here power of the expansion,

$n=11$

Degree

$=n+1=11+1=12$

n is odd so two middle terms will be, i.e

$(\dfrac{(n+1)}{2})=6$ th term and

$(\dfrac{(n+1)}{2})+1)=7$ th term

$T_{6}=\binom{11}{5}\times \left ( x^{2} \right )^{6}\times \left ( \frac{1}{x} \right )^{5}$

$T_{7}=\binom{11}{6}\times \left ( x^{2} \right )^{5}\times \left ( \frac{1}{x} \right )^{6}$

Prove that: $\sum_{r \, = \, 0}^{n} \, r(n \, - \, r)C_r^2 \, = \, n^2(^{2n \, - \, 2}C_n)$

$(1 \, + \, x)^n \, = \, C_0 \, + \, C_1 \, x \, + \, C_2 \, x^2 \, + \, ...... \, + \, C_r \, x^r \, + \, ..... \, + C_n \, x^n$ .....(1)
Now the term r(n - r)

$C_r^2$ suggests the factor
r(

$C_r$ ) (n - r)

$C_r$
By complementary rule

$C_r \, = \, C_{n \, - \, r}$
Against r

$C_r$ suggest differentiation. The second factor suggest that (1) should be written as

$(x \, + \, 1)^n \, = \, C_0x^n \, + \, C_1 x^{n \, - \, 1} \, + \, ..... \, + \, C_r x^{n \, - \, r} \, + \, .... \, + \, C_n$ .....(2)
and the second factor (n - r)

$C_r$ suggest again differentiation of (2).
Differentiate both (1) and (2)

$n (1 \, + \, x)^{n \, - \, 1} \, = \, C_1 \, + \, 2 C_2 x \, + \, ...... \, + \, rC_r x^{r - 1} \, + \, .....$ ....(3)

$n (1 \, + \, x)^{n \, - \, 1} \, = \, n C_0 x^{n \, - \, 1} \, + \, (n \, - \, 1)C_1 x^{n \, - \, 2} \, + \, ...... \, + \, (n \, - \, r)C_r x^{n \, - \, r \, - \, 1} \, + \, ......$ ...(4)
Multiplying (3) and (4), we get

$n^2 \, (1 \, + \, x)^{2n \, - \, 2} \, = \, \sum \, (rC_r) \, (n \, - \, r) \, C_r x^{r \, - \, 1} \, x^{n \, - \, r \, - \, 1}$

$= \, \sum \, r \, (n \, - \, r) \, C_r^2 \, x^{n \, - \, 2}$
Hence we should search the coefficient of

$x^{n \, - \, 2}$ in L.H.S. and it will be

$n^2 \, \cdot \, ^{2n \, - \, 2}C_{n \, - \, 2} \, = \, n^2 \, (^{2n \, - \, 2}C_n)$ by complementary rule

If $(1+x)^n=C_0+C_1x+C_2x^2+...+C_nx^n$ , then the value of $C_0+C_2+C_4+....$ is?

1329058_1084199_ans_058384c290264146af3af60ef7b63606.jpg

If ${\left(x+\dfrac{1}{x}+1\right)}^{6}={a}_{0}+\left({a}_{1}x+\dfrac{{b}_{1}}{x}\right)+\left({a}_{2}{x}^{2}+\dfrac{{b}_{2}}{{x}^{2}}\right)+...+\left({a}_{6}{x}^{6}+\dfrac{{b}_{6}}{{x}^{6}}\right)$ then find the value of ${a}_{0}$

$\because {\left(x+\dfrac{1}{x}+1\right)}^{6}=\sum_{r=0}^{r=6}{^{6}{{C}_{r}}{\left(x+\dfrac{1}{x}\right)}^{r}}$ for constant term $r$ must be even integer.

$\therefore {a}_{0}=^{6}{{C}_{0}}+^{6}{{C}_{2}}\times ^{6}{{C}_{1}}+^{6}{{C}_{4}}\times ^{6}{{C}_{2}}+^{6}{{C}_{6}}\times ^{6}{{C}_{3}}$

$=1+\dfrac{6\times 5\times 4!}{4!\times 2}\times \dfrac{6\times 5!}{5!\times 1}+\dfrac{6\times 5\times 4!}{4!\times 2}\times \dfrac{6\times 5\times 4!}{4!\times 2}+1\times \dfrac{6\times 5\times 4\times 3!}{3!3!}$

$=1+30+90+20=141$

Find the integral part of $(3 + \sqrt {7})^{5}$ .

1353067_1130155_ans_998c0c89d5714461827d497da8b3e454.jpg

$Let\,\,n \in N;{S_n} = \sum\limits_{r = 0}^{3n} {\left( {^{3n}{C_{3r}}} \right)} .\,\,Find\,\,\left| {{S_n} - 3{T_n}} \right|.$

2041610_1059499_ans_8974616fcb1e497db2d0057d9cb746ab.jpg

If log 1001= 3.000434, find the number of digits in ${1001^{101}}$

given:

$\Rightarrow \ \log 1001=3.000434$

multiplying both side by

$101$

$\Rightarrow \ 101 \log 1001=101 \times 3.000434$

$\Rightarrow \ \log 1001^{101}=101\times 3.000434$

$\Rightarrow \ 1001^{101}=10^{101\times 3.000434}$

$\Rightarrow \ 1001^{101}=10^{303.043834}$

(as standard base of

$\log$ is

$10$ )

$\therefore \$ No. of digit should be greater than

$303$

Find the sum of coefficient of the expression
${(x+2y+4z)}^{10}$

for sum of coefficients

$x=y=z=1$

$=(1+2+4)^{10}$

$=7^{10}$

In the expansion of ${ \left( { 7 }^{ 1/3 }+{ 11 }^{ 1/9 } \right) }^{ 6561 }$ , prove that there will be only $730$ terms which are free from radicals.

General term :-

$^{6561}C_r \, 7^{\dfrac{1}{3} (6561 - r)} \, 11^{r/9}$

For terms to be free of radicals

$r = 0, 9, 18 ,..., 6561$

it forms an AP

$An = a + (n - 1) d$

$6561 = 0 + (n - 1) 9$

$n - 1 = \dfrac{6561}{9} = 729$

$n = 730$

$\therefore$ there will only be

$730$ terms which are free form radicals

Find the term containing $x^{2}$ , if any, in the expansion of $(3x-\dfrac{1}{2x})^{8}$

${ \left( 3x-\dfrac { 1 }{ 2x } \right) }^{ 8 }\Rightarrow { \left( a+b \right) }^{ n }=\sum _{ r=0 }^{ n }{ \left( \begin{matrix} n r \end{matrix} \right) } { a }^{ n-r }{ b }^{ r }$

${ T }_{ r+1 }=^{ n }{ C }_{ r }{ a }^{ n-r }{ b }^{ r }$

${ T }_{ r+1 }=^{ 8 }{ C }_{ r }\left( { 3x } \right) ^{ 8-r }{ \left( \dfrac { 1 }{ 2x } \right) }^{ r }$

$8-r-r=2$

$2r=6$

$\boxed{r=3}$

${ T }_{ 4 }=^{ 8 }{ C }_{ r }\left( { 3x } \right) ^{ 5 }{ \left( \dfrac { -1 }{ 2x } \right) }^{ 3 }$

The coefficient of $x^{4}$ in the expansion of $(\frac{x}{2}-\frac{3}{x^{2}})^{10}$ is equal to:

We have,

${{\left( \dfrac{x}{2}-\dfrac{3}{{{x}^{2}}} \right)}^{10}}$

We know that, term of ${{x}^{r}}$ in the ${{\left( x+a \right)}^{n}}$ is

${{\left( x+a \right)}^{n}}=\sum\limits_{r=0}^{n}{^{n}{{C}_{r}}{{x}^{n-r}}{{a}^{r}}}$

Then,

${{\left( \dfrac{x}{2}-3{{x}^{-2}} \right)}^{10}}=\sum\limits_{k=0}^{10}{^{10}{{C}_{k}}{{\left( \dfrac{x}{2} \right)}^{10-k}}{{\left( -3{{x}^{-2}} \right)}^{k}}}$

$=\sum\limits_{k=0}^{10}{^{10}{{C}_{k}}\dfrac{{{x}^{10-k}}}{{{2}^{^{10-k}}}}\left( -{{3}^{k}}{{x}^{-2k}} \right)}$

$=-\sum\limits_{k=0}^{10}{^{10}{{C}_{k}}{{x}^{10-k}}{{2}^{k-10}}{{3}^{k}}{{x}^{-2k}}}$

$=-\sum\limits_{k=0}^{10}{^{10}{{C}_{k}}{{x}^{10-3k}}{{2}^{k-10}}{{3}^{k}}}$

We then observe that, the ${{x}^{4}}$ term coincides with $k=2$

Because,

$={{-}^{10}}{{C}_{2}}{{x}^{10-3\times 2}}{{2}^{2-10}}{{3}^{2}}$

$={{-}^{10}}{{C}_{2}}\times {{2}^{-8}}\times {{3}^{2}}{{x}^{4}}$

$=-\dfrac{10!}{2!\left( 10-2 \right)!}\times {{2}^{-8}}\times {{3}^{2}}{{x}^{4}}$

$=-\dfrac{10\times 9\times 8!}{2\times 1\times 8!}\times {{2}^{-8}}\times {{3}^{2}}{{x}^{4}}$

$=-45\times {{2}^{-8}}\times {{3}^{2}}{{x}^{4}}$

$=-45\times 9\times {{2}^{-8}}{{x}^{4}}$

$=-\dfrac{45\times 9}{{{2}^{8}}}{{x}^{4}}$

$=-\dfrac{405}{256}{{x}^{4}}$

Hence, the coefficient of ${{x}^{4}}$ is $-\dfrac{405}{256}$ .

Prove that
$C_{1}+2 C_{2}+3 C_{3}+4 C_{4}+.+ ^nC_{n}=n.2^{n-1}$

$C_1+2C_2+3C_3+....+nC_n$

$=n+\dfrac{2n(n-1)}{2!}+\dfrac{3n(n-1)(n-2)}{3!}+.....+n\times 1$

$=n+n(n-1)+\dfrac{n(n-1)(n-2)}{2}+......+n$

$=n\left[1+n-1+\dfrac{(n-1)(n-2)}{2}+.....+1\right]$

$=n\left[1+\dfrac{n-1}{1!}+\dfrac{(n-1)(n-2)}{2}+.....+1\right]$

put

$n-1=N$

$=n\left[1+\dfrac{N}{1!}+\dfrac{N(N+1)}{2!}+.....+1\right]$

$=n\left[NC_0+NC_1+NC_2+......+NC_N\right]$

$=n 2^N$

$=n2^{n-1}$ Hence proved.

In the expansion of ${\left({3}^{\tfrac{-x}{4}}+{3}^{\tfrac{5x}{4}}\right)}^{n},$ if the sum of the binomial coefficeints is $64$ and the term with the greatest binomial coefficient exceeds the third by $\left(n-1\right)$ ,then the value of $x$ is

1339880_1232708_ans_a07a77e539f2430489d670695727a15f.PNG

Show that, if the greatest term in the expansion of ${\left(1+x\right)}^{2n}$ has also the greatest coefficient , then $x$ lies between $\dfrac{n}{n+1}$ and $\dfrac{n+1}{n}$

1339877_1232537_ans_807ef7dc8ffd47c588eb8b7fada2a3fd.PNG

The numerically greatest term in the expansion of ${\left(2+3x\right)}^{9}$ when $x=\dfrac{3}{2}$ is

1339875_1232523_ans_743db7513a334e0fb505637e282ee61e.PNG

Coefficient of ${x}^{7}$ in the expansion of $\left( {1+3x-{2x}^{3}} \right )^{10}$

2039334_1262578_ans_02592fe8ecd743ba80e34179f918459c.JPG

Find $a$ if the coefficients of ${x}^{2}$ and ${x}^{3}$ in the expansion of ${\left(3+ax\right)}^{3}$ are equal.

We know that

General term of expansion

${\left(a+b\right)}^{n}$ is

${T}_{r+1}=^{n}C_{r}{a}^{n-r}{b}^{r}$

For

${\left(3+ax\right)}^{9}$ ,

Put

$a=3,b=9x$ and

$n=9$

General term of

${\left(3+ax\right)}^{9}$ is

${T}_{r+1}=^{9}C_{r}{3}^{9-r}{\left(ax\right)}^{r}=^{9}C_{r}{3}^{9-r}{a}^{r}{x}^{r}$ ......

$(1)$

We need to find the coefficient of

${x}^{2}$ and

${x}^{3}$

Put

$r=2$ in

$(1)$

${T}_{2+1}=^{9}C_{2}{3}^{9-2}{a}^{2}{x}^{2}$

$=^{9}C_{2}{3}^{7}{a}^{2}{x}^{2}$

Coefficient of

${x}^{2}$ is

$^{9}C_{2}{3}^{7}{a}^{2}$

Put

$r=3$ in

$(1)$

${T}_{3+1}=^{9}C_{3}{3}^{9-3}{a}^{3}{x}^{3}$

$=^{9}C_{3}{3}^{6}{a}^{3}{x}^{3}$

Coefficient of

${x}^{2}$ is

$^{9}C_{3}{3}^{6}{a}^{3}$

Given that Coefficient of

${x}^{2}=$ Coefficient of

${x}^{3}$

$\Rightarrow ^{9}C_{2}{3}^{7}{a}^{2}=^{9}C_{3}{3}^{9-3}{a}^{3}$

$\Rightarrow \dfrac{9!}{2!\left(9-2\right)!}{3}^{7}{a}^{2}=\dfrac{9!}{3!\left(9-3\right)!}{3}^{6}{a}^{3}$

$\Rightarrow \dfrac{9!}{2!7!}{3}^{7}{a}^{2}=\dfrac{9!}{3!6!}{3}^{6}{a}^{3}$

$\Rightarrow a=\dfrac{3\times 2!\times 6!}{2!\times 7\times 6!}\times3$

$\therefore a=\dfrac{9}{7}$

Find the coefficient of $x^5$ in the product ${\left(1+2x\right)}^{6}{\left(1-x\right)}^{7}$ using binomial theorem.

${\left(1+2x\right)}^{6}={{6}_{C}}_{0}+{{6}_{C}}_{1}2x+{{6}_{C}}_{2}{\left(2x\right)}^{2}+{{6}_{C}}_{3}{\left(2x\right)}^{3}+{{6}_{C}}_{4}{\left(2x\right)}^{4}+{{6}_{C}}_{5}{\left(2x\right)}^{5}+{{6}_{C}}_{6}{\left(2x\right)}^{6}$

$=1+6\times 2x+15\times {\left(2x\right)}^{2}+20\times {\left(2x\right)}^{3}+15\times {\left(2x\right)}^{4}+6\times {\left(2x\right)}^{5}+{\left(2x\right)}^{6}$

$=1+12x+60{x}^{2}+160{x}^{3}+240{x}^{4}+192{x}^{5}+64{x}^{6}$

${\left(1-x\right)}^{7}={{7}_{C}}_{0}-{{7}_{C}}_{1}x+{{7}_{C}}_{2}{x}^{2}-{{7}_{C}}_{3}{x}^{3}+{{7}_{C}}_{4}{x}^{4}-{{7}_{C}}_{5}{x}^{5}+{{7}_{C}}_{6}{x}^{6}-{{7}_{C}}_{7}{x}^{7}$

$=1-7x+21{x}^{2}-35{x}^{3}+35{x}^{4}-21{x}^{5}+7{x}^{6}-{x}^{7}$

$\therefore{\left(1+2x\right)}^{6}{\left(1-x\right)}^{7}=\left(1+12x+60{x}^{2}+160{x}^{3}+240{x}^{4}+192{x}^{5}+64{x}^{6}\right)\\\left(1-7x+21{x}^{2}-35{x}^{3}+35{x}^{4}-21{x}^{5}+7{x}^{6}-{x}^{7}\right)$

The complete multiplication of the two bracets is not required to be carried out.Only those terms, which involve

${x}^{5}$ are required.

The terms containing

${x}^{5}$ are

$=1\left(-21x^5\right)+\left(12x\right)\left(35{x}^{4}\right)+\left(60{x}^{2}\right)\left(-35{x}^{3}\right)+\left(160{x}^{3}\right)\left(21{x}^{2}\right)+\left(240{x}^{4}\right)\left(-7x\right)+\left(192{x}^{5}\right)\left(1\right)$

$=171{x}^{5}$ on simplification

Thus, the coefficient of

${x}^{5}$ in the given product is

$171$

Find $a, b$ and $n$ in the expression of $(a+b)^n$ if the first three terms of the expansion are $729, 7290$ and $30375$ respectively.

Its known that

${\left(r+1\right)}^{th}$ term,

${T}_{r+1}$ in the binomial expansion of

${\left(a+b\right)}^{n}$ is

${T}_{r+1}={ { n }_{ C } }_{ r }{a}^{n-r}{b}^{r}$

The first three terms of the expansion are given as

$729,7290,$ and

$30375$ respectively.

$\therefore$ we obtain

${T}_{1}={{n}_{C}}_{0}{a}^{n-0}{b}^{0}\Rightarrow {a}^{n}=729$ ......

$\left(1\right)$

${T}_{2}={{n}_{C}}_{1}{a}^{n-1}{b}^{1}\Rightarrow n{a}^{n-1}b=7290$ ......

$\left(2\right)$

${T}_{3}={{n}_{C}}_{2}{a}^{n-2}{b}^{2}\Rightarrow \dfrac{n\left(n-1\right)}{2}{a}^{n-2}{b}^{2}=30375$ ......

$\left(3\right)$

Dividing

$\left(2\right)$ by

$\left(1\right)$

$\Rightarrow \dfrac{n{a}^{n-1}b}{{a}^{n}}=\dfrac{7290}{729}=10$

$\Rightarrow \dfrac{nb}{a}=10$ .....

$\left(4\right)$

Dividing

$\left(3\right)$ by

$\left(2\right)$

$\Rightarrow \dfrac{n\left(n-1\right){a}^{n-2}{b}^{2}}{2n{a}^{n-1}b}=\dfrac{30375}{7290}$

$\Rightarrow \dfrac{\left(n-1\right)b}{a}=\dfrac{30375}{7290}\times 2=\dfrac{25}{3}$

$\Rightarrow \dfrac{nb}{a}-\dfrac{b}{a}=\dfrac{25}{3}$

$\Rightarrow \dfrac{b}{a}=10-\dfrac{25}{3}=\dfrac{30-25}{3}=\dfrac{5}{3}$ .......

$\left(5\right)$

From

$\left(4\right)$ and

$\left(5\right)$ we get

$n.\dfrac{b}{a}=n.\dfrac{5}{3}=10$

$\Rightarrow n=6$ on simplification

Substituting

$n=6$ in eqn

$\left(1\right)$ we get

${a}^{6}=729$

$\therefore a={729}^{\frac{1}{6}}=3$

From eqn

$\left(5\right)$ we obtain

$\dfrac{b}{3}=\dfrac{5}{3}\Rightarrow b=5$

Thus,

$a=3,b=5$ and

$n=6$

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left( \sqrt [ 4 ] { 2 } + \frac { 1 } { \sqrt [ 4 ] { 3 } } \right) ^ { ( n ) }$ is $\sqrt { 6 } : 1$

2041259_1457952_ans_472d324426274d39813748d3852c0a76.jpg

Show that the middle term in the expansion of $(x - \frac{1}{x})^2n$ is $\frac{1.3.5.....(2n-1)}{n;} (-2)^n$

2044400_1564248_ans_329e1946c4de46a6a46e7cbe8c03a826.jpg

Find the coefficient of $x^5$ in the expansion of $( 1+ x)^3(1-x)^6$ .

Given expression is,

${ \left( 1+x \right) }^{ 3 }{ \left( 1-x \right) }^{ 6 }$

$={ \left( 1+x \right) }^{ 3 }{ \left( 1-x \right) }^{ 3 }{ \left( 1-x \right) }^{ 3 }$

$=\left[ { \left( 1+x \right) }\left( 1-x \right) \right] ^{ 3 }{ \left( 1-x \right) }^{ 3 }$

$=\left[ 1-{ x }^{ 2 } \right] ^{ 3 }{ \left( 1-x \right) }^{ 3 }$

We know that

${ \left( a-b \right) }^{ 3 }={ a }^{ 3 }-3{ a }^{ 2 }b+3ab^{ 2 }-{ b }^{ 3 }$

$\therefore \left[ 1-{ x }^{ 2 } \right] ^{ 3 }{ \left( 1-x \right) }^{ 3 }=\left[ { \left( 1 \right) }^{ 3 }-3{ \left( 1 \right) }^{ 2 }{ x }^{ 2 }+3{ \left( 1 \right) }{ \left( { x }^{ 2 } \right) }^{ 2 }-{ \left( { x }^{ 2 } \right) }^{ 3 } \right] \left[ { \left( 1 \right) }^{ 3 }-3{ \left( 1 \right) }^{ 2 }x-3{ \left( 1 \right) { x }^{ 2 } }-{ x }^{ 3 } \right]$

$=\left[ 1-3{ x }^{ 2 }+3{ x }^{ 4 }-{ x }^{ 6 } \right] \left[ 1-3x+3{ { x }^{ 2 } }-{ x }^{ 3 } \right]$

$=1-3x+3{ { x }^{ 2 } }-{ x }^{ 3 }-3{ { x }^{ 2 } }+6{ x }^{ 3 }-9{ x }^{ 4 }+3{ x }^{ 5 }+3{ x }^{ 4 }-9{ x }^{ 5 }+9{ x }^{ 6 }-3{ x }^{ 7 }-{ x }^{ 6 }+3{ x }^{ 7 }-3{ x }^{ 8 }+{ x }^{ 9 }$

$=1-3x+5{ x }^{ 3 }-6{ x }^{ 4 }-6{ x }^{ 5 }+8-6{ x }^{ 6 }-3{ x }^{ 8 }+{ x }^{ 9 }$

Thus, coefficient of

${ x }^{ 5 }$ is -6

Show that the coefficient of $x^4$ in the expansion of $( \frac {x}{2} - \frac {3}{x^2} )^{10}$ is $\frac {405}{256}$

1848741_1709597_ans_dceae93fad434c489f3fecc56f796efa.jpeg

If the 17th and 18th terms in the expansion of $( 2+a )^{50}$ are equal , find the value of a.

nth term of binomial expansion

${ \left( a+b \right) }^{ m }$ is given by,

${ a }_{ n }=\left( \begin{matrix} m \\ n-1 \end{matrix} \right) { a }^{ m+1-n }b^{ n-1 }$

In given problem,

$a=2$ ,

$b=a$ ,

$m=50$ ,

${ n }_{ 1 }=17$ and

${ n }_{ 2 }=18$

As given in problem, 17th and 18th terms are equal.

$\therefore \left( \begin{matrix} 50 \\ 17-1 \end{matrix} \right) { \left( 2 \right) }^{ 50+1-17 }{ a }^{ 17-1 }=\left( \begin{matrix} 50 \\ 18-1 \end{matrix} \right) { \left( 2 \right) }^{ 50+1-18 }{ a }^{ 18-1 }$

$\therefore \left( \begin{matrix} 50 \\ 16 \end{matrix} \right) { \left( 2 \right) }^{ 34 }{ a }^{ 16 }=\left( \begin{matrix} 50 \\ 17 \end{matrix} \right) { \left( 2 \right) }^{ 33 }{ a }^{ 17 }$

$\therefore \frac { 50! }{ 16!\left( 50-16 \right) ! } { \left( 2 \right) }^{ 34 }{ a }^{ 16 }=\frac { 50! }{ 17!\left( 50-17 \right) ! } { \left( 2 \right) }^{ 33 }{ a }^{ 17 }$

$\therefore \frac { 50! }{ 16!\times 34\times 33! } \times 2\times { \left( 2 \right) }^{ 33 }{ a }^{ 16 }=\frac { 50! }{ 17\times 16!\times 33! } { \left( 2 \right) }^{ 33 }\times a\times { a }^{ 16 }$

$\therefore \frac { 2 }{ 34 } =\frac { a }{ 17 }$

$\therefore 34a=34$

$\therefore a=1$

Find the 9th term in the expansion of $( \frac {a}{b} - \frac {b}{2a^2})^{12}$

Formula to find out

${ n }^{ th }$ term in binomial expansion of

${ \left( p+q \right) }^{ m }$

${ n }^{ th }\quad term=\left( \begin{matrix} ^m c_{n-1}\\ \end{matrix} \right) { p }^{ m+1-n }{ q }^{ n-1 }$

where,

$\left( \begin{matrix} ^m c_{n-1} \end{matrix} \right) =\dfrac { m! }{ n!\left( m-n \right) ! }$

For given problem,

$p=\dfrac { a }{ b }$ ,

$q=\dfrac { -b }{ 2{ a }^{ 2 } }$ ,

$m=12$

we have to find

${ 9 }^{ th }$ term. Thus,

$n=9$

$\therefore { 9 }^{ th }term=\left( \begin{matrix} ^{12} c_{8} \end{matrix} \right) { \left( \dfrac { a }{ b } \right) }^{ 12+1-9 }{ \left( \dfrac { -b }{ 2{ a }^{ 2 } } \right) }^{ 9-1 }$

$\therefore { 9 }^{ th }term=\dfrac { 12! }{ 8!\left( 12-8 \right) ! } { \left( \frac { a }{ b } \right) }^{ 12+1-9 }{ \left( \frac { -b }{ 2{ a }^{ 2 } } \right) }^{ 9-1 }$

$\therefore { 9 }^{ th }term=\dfrac { 12! }{ 8!\times 4! } { \left( \frac { a }{ b } \right) }^{ 12+1-9 }{ \left( \frac { -b }{ 2{ a }^{ 2 } } \right) }^{ 9-1 }$

$\therefore { 9 }^{ th }term=\dfrac { 12\times 11\times 10\times 9\times 8! }{ 8!\times 4! } { \left( \frac { a }{ b } \right) }^{ 4 }{ \left( \frac { -b }{ 2{ a }^{ 2 } } \right) }^{ 8 }$

$\therefore { 9 }^{ th }term=\dfrac { 12\times 11\times 10\times 9 }{ 4\times 3\times 2 } \left( \dfrac { { a }^{ 4 } }{ b^{ 4 } } \right) \left( \dfrac { b^{ 8 } }{ 256{ a }^{ 16 } } \right)$

$\therefore { 9 }^{ th }term=\dfrac { 495 }{ 256 } \left( \dfrac { b^{ 4 } }{ { a }^{ 12 } } \right)$

Find the term independent of $x$ in the expansion of the following expressions:
$(1+x+2x^{3})\left(\dfrac{3}{2}x^{2}-\dfrac{1}{3x}\right)^{9}$

Enter 1 if answer is $\dfrac {17}{54}$ otherwise enter 0.

1414669_1666828_ans_84978291b49d41e0989261119da4893b.jpg

If the middle term in the expansion of $( \frac {p}{2} + 2 )^8$ is 1120, find p.

1947901_1709575_ans_3a6f387e058f401d9a07a15571feb5b4.jpg

Find the sixth term in the expansion $(y^{1/2}+x^{1/3})^{n}$ , if the binomial coefficient of the third term from the ends is $45$ .

In the binomial expansion of

$\left(y^\dfrac{1}{2}+x^\dfrac{1}{3}\right)^{n}$ ,there are

$\left(n+1\right)$ terms.

The third term from the end is

$\left(\left(n+1\right)-3+1\right)^{th}$ i.e

$\left(n-1\right)^{th}$ term from the beginning.

$\therefore$ The binomial coefficient of

$3rd$ term from the end

$=$ The binomial coefficient of

$\left(n-1\right)^{th}$ term from the beginning

$=^{n}{{C}_{n-2}}=^{n}{{C}_{2}}$

It is given that the binomial coefficient of the third term from the end is

$45$ .

$\therefore^{n}{{C}_{2}}\Rightarrow\dfrac{n\left(n-1\right)}{2}=45 \Rightarrow n^{2}-n-90\Rightarrow\left(n-10\right)\left(n+9\right)=0\Rightarrow n=10.$

Let

$T_{6}$ be the sixth term in the binomial expansion of

$\left(y^{{1}/{2}}+x^{{1}/{3}}\right)^{n}$ .Then,

$T_{6}=^{n}{{C}_{5}} \left(y^{\dfrac{1}{2}}\right)^{n-5}$

$\left(x^{{1}/{3}}\right)^{5}=^{10}{{C}_{5}}y^{{5}/{2}}x^{{5}/{3}}=252y^{{5}/{2}} x^{{5}/{3}}$

Find the 6th term if expansion $( y^{1/2} + x^{ 1/3} )^n$ , if the binominal coefficient of 3rd term form the end is 45.

We know that nth term of binomial expansion

${ \left( a+b \right) }^{ m }$ is given by,

${ a }_{ n }=\left( \begin{matrix} m \\ n-1 \end{matrix} \right) { \left( a \right) }^{ m+1-n }{ \left( b \right) }^{ n-1 }$

3rd term from the last will be (n-1)th term.

Thus, in above equation,

$n=n-1$

$a={ y }^{ \frac { 1 }{ 2 } }$

$b={ x }^{ \frac { 1 }{ 3 } }$

$m=n$

$\therefore { a }_{ n-1 }=\left( \begin{matrix} n \\ \left( n-1 \right) -1 \end{matrix} \right) { \left( { y }^{ \frac { 1 }{ 2 } } \right) }^{ n+1-\left( n-1 \right) }{ \left( { x }^{ \frac { 1 }{ 2 } } \right) }^{ \left( n-1 \right) -1 }$

$\therefore { a }_{ n-1 }=\left( \begin{matrix} n \\ n-2 \end{matrix} \right) { \left( { y }^{ \frac { 1 }{ 2 } } \right) }^{ 2 }{ \left( { x }^{ \frac { 1 }{ 2 } } \right) }^{ n-2 }$

$\therefore { a }_{ n-1 }=\frac { n! }{ \left( n-2 \right) !\left[ n-\left( n-2 \right) \right] ! } y{ \left( { x } \right) }^{ \frac { n-2 }{ 2 } }$

As given in the problem,

$\frac { n! }{ \left( n-2 \right) !\left[ n-\left( n-2 \right) \right] ! } =45$

$\therefore \frac { n\left( n-1 \right) \left( n-2 \right) ! }{ \left( n-2 \right) !\times 2! } =45$

$\therefore \frac { n\left( n-1 \right) }{ 2 } =45$

$\therefore n\left( n-1 \right) =90$

$\therefore { n }^{ 2 }-n-90=0$

$\therefore n=\frac { -\left( -1 \right) \pm \sqrt { { \left( -1 \right) }^{ 2 }-4\times 1\times \left( -90 \right) } }{ 2\times 1 }$

$\therefore n=\frac { 1\pm \sqrt { 1+360 } }{ 2 }$

$\therefore n=\frac { 1\pm \sqrt { 361 } }{ 2 }$

Consider only positive root.

$\therefore n=\frac { 1+19 }{ 2 }$

$\therefore n=\frac { 20 }{ 2 } =10$

Thus, expression will become,

${ \left( { y }^{ \frac { 1 }{ 2 } }+x^{ \frac { 1 }{ 3 } } \right) }^{ 10 }$

Now, 6th term will be,

${ a }_{ 6 }=\left( \begin{matrix} 10 \\ 6-1 \end{matrix} \right) { \left( y^{ \frac { 1 }{ 2 } } \right) }^{ 10+1-6 }{ \left( x^{ \frac { 1 }{ 3 } } \right) }^{ 6-1 }$

$\therefore { a }_{ 6 }=\left( \begin{matrix} 10 \\ 5 \end{matrix} \right) { \left( y^{ \frac { 1 }{ 2 } } \right) }^{ 5 }{ \left( x^{ \frac { 1 }{ 3 } } \right) }^{ 5 }$

$\therefore { a }_{ 6 }=\frac { 10! }{ 5!\left( 10-5 \right) ! } { \left( y \right) }^{ \frac { 5 }{ 2 } }{ \left( x \right) }^{ \frac { 5 }{ 3 } }$

$\therefore { a }_{ 6 }=\frac { 10! }{ 5!\times 5! } { \left( y \right) }^{ \frac { 5 }{ 2 } }{ \left( x \right) }^{ \frac { 5 }{ 3 } }$

$\therefore { a }_{ 6 }=\frac { 10\times 9\times 8\times 7\times 6\times 5! }{ 5!\times 5! } { \left( y \right) }^{ \frac { 5 }{ 2 } }{ \left( x \right) }^{ \frac { 5 }{ 3 } }$

$\therefore { a }_{ 6 }=\frac { 10\times 9\times 8\times 7\times 6 }{ 120 } { \left( y \right) }^{ \frac { 5 }{ 2 } }{ \left( x \right) }^{ \frac { 5 }{ 3 } }$

$\therefore { a }_{ 6 }=252{ \left( y \right) }^{ \frac { 5 }{ 2 } }{ \left( x \right) }^{ \frac { 5 }{ 3 } }$

Show that the middle term in the expansion of $( \frac {2x^3}{3} + \frac {3}{2x^2} )^{10}$ is 252

1947902_1709596_ans_7c800b5ec01448c3910e0aa3a209499d.jpg

Show that the coefficient of $x^{-3}$ in the expansion of $( x -\frac {1}{x} )^{11}$ is -330

1848738_1709595_ans_7a731a8e10df43259533097dd0ce8f8b.jpeg

Write the number of terms in the expansion of $(\sqrt {2} + 1)^5 +( \sqrt {2} -1)^5$ .

1855643_1709602_ans_08f55a3319bb41cfa2e990fe1977daa8.jpeg

If the coefficient of $(r-5)^{th}$ and $(2r -1)^{th}$ terms in the expansion of $(1+ x)^{34}$ are equal, find the value of $r.$

1854981_1709607_ans_bbe4753ae27443e9a1ff14a5fdf202c7.jpeg

Write the coefficient of $x^7y^2$ in the expansion of $(x +2y)^9$ .

1948992_1709608_ans_5fe3f1dbaa134d5d9d45563de69668c6.jpg

Prove that there is no term involving $x^6$ in the expansion $(2x^2 - \frac {3}{x})^{11}$ .

1848742_1709598_ans_d9ba61ee4f0b4d9591400c964e361034.jpeg

Show that the coefficient of $x^4$ in the expansion of $( 1+ 2x+ x^2)^5$ is 210.

1948987_1709600_ans_29606a92354d4c8f997e2abdb738bcc1.jpg

Write the coefficient of the middle term in the expansion of $( 1 +x)^{2n} .$

1948990_1709604_ans_d3d1911fc23c4a49b99c6a7d3e591583.jpg

Write down and simplify:
The $4^{th}$ term of $(x-5)^{13}$ .

1972363_1743498_ans_a7dc7c1e54b644459f784a0a7269c2d2.jpg

Write down and simplify:
The $12^{th}$ term of $(2x-1)^{13}$ .

1972364_1743501_ans_5c778caf2c7d441c800290fa2206ebd2.jpg

Write down and simplify:
The $5^{th}$ term of $\left(\frac{x^{\frac{3}{2}}}{a^{\frac{1}{2}}}-\frac{y^{\frac{5}{2}}}{b^{\frac{3}{2}}}\right)^8$ .

1973027_1743506_ans_b67a882935fe48b690d1609a556592e7.jpg

Write down and simplify:
The $5^{th}$ term of $\left(2a-\dfrac{b}{3}\right)^8$ .

2037858_1743504_ans_01a9105086524263979a85ab82650448.jpg

Find the $13^{th}$ term of $\left(9x-\dfrac{1}{3\sqrt{x}}\right)^{18}$ .

1885354_1743510_ans_1fcb455c25f242d6bddaa53478e0c1eb.jpeg

Find the middle term of $\left(\dfrac{a}{x}+\dfrac{x}{a}\right)^{10}$ .

1973337_1743508_ans_8bc15bcb2e214ae0b586b4b9183073a7.jpg

Write down and simplify:
The $7^{th}$ term of $\left(\dfrac{4x}{5}-\dfrac{5}{2x}\right)^9$ .

1885345_1743505_ans_4bf0a1a3ac71406e8949de861c6486ed.jpeg

Write down and simplify:
The $10^{th}$ term of $(1-2x)^{13}$ .

1972362_1743500_ans_a1cf30ab5249444e9215eea3c2f4a2db.jpg

Write down and simplify:
The $28^{th}$ term of $(5x+8y)^{30}$ .

1885336_1743502_ans_1788e8f9086a49fbba302190b3d19863.jpeg

Write down and simplify:
The $4^{th}$ term of $\left(\dfrac{a}{3}+9b\right)^{10}$ .

1973031_1743503_ans_ba5418811fbc4c0a899269269da1b9ea.jpg

Prove that the coefficient of $x^n$ in the expansion $\dfrac{1}{1 + x+ x^2}$ is $1, 0, -1$ according as $n$ is of the form $3m, \ 3m-1, or \ 3m+1$ .

1981370_1743873_ans_439a6f1bc55d46da861f9b558659d8b5.jfif

Binomial Theorem - Class 11 Engineering Maths - Extra Questions

Find the middle terms in the expansion of (5x−7y)7(5x-7y)^7.

Find the 28th term of (5x+8y)30(5x+8y)^{30}.

Show that the expansion of (x2+1x)12\left( x^2 + \dfrac{1}{x} \right)^{12} does not contain any term involving x−1x^{-1}.

Find the middle term in the expansion (xa+yb)6{ \left( \cfrac { x }{ a } +\cfrac { y }{ b } \right) }^{ 6 }.

Write down and simplify:The 4th term of (x−5)13{ \left( x-5 \right) }^{ 13 }

Find the 13th13^{th} term in the expansion of (9x−13√x)18{ \left( 9x-\cfrac { 1 }{ 3\sqrt { x } } \right) }^{18 }.

The sum of the coefficients in the first three terms of the expansion of (x2−2x)m\displaystyle\, \left ( x^2 - \frac{2}{x} \right )^m is equal to 9797. Find the term of the expansion containing x4x^4.

Write general term of this:-2x(3+2x2)202x{\left( {3 + 2{x^2}} \right)^{20}}

Find the middle term of (ax+xa)10{ \left( \cfrac { a }{ x } +\cfrac { x }{ a } \right) }^{ 10 }.

Show that nC0− nC1+nC2 −.....=0\displaystyle ^{ n }C_{ 0 } -\ ^{ n }C_{ 1 }+^{ n }C_{ 2 }\ -.....=0

Write down and simplify:The 7th term of (4x5−52x)9{ \left( \cfrac { 4x }{ 5 } -\cfrac { 5 }{ 2x } \right) }^{ 9 }\quad

Find the sum n∑r=1rnCrnCr−1\displaystyle \sum_{r=1}^{n}\dfrac {r {^{n}C_{r}}}{{^{n}C_{r-1}}}.

Find the 7th7 ^ { th } term of (3x2−13)10\left( 3 x ^ { 2 } - \frac { 1 } { 3 } \right) ^ { 10 }.

If 4th4^{th} term in the expansion of (ax+1x)n\left(ax+\dfrac{1}{x}\right)^n is 52\dfrac{5}{2}, then find the value of aa and nn.

Find the coefficient of x5x^5 in the expansion of the product (1+2x)6(1−x)7(1+2x)^6(1-x)^7.

15C3+15C5+.......+15C15^{ 15 }{ C_{ 3 } }{ +^{ 15 } }{ C_{ 5 } }+.......{ +^{ 15 } }{ C_{ 15 } }

Show that the middle term in the expansion of (1+x)n(1 + x)^{n} is 6x2 6x^2 if n=4n=4

The sum of the rational terms in the expansion of (√2+31/5)10(\sqrt{2}+3^{1/5})^{10} is _______.

The sum of the coefficient of the polynomial (1+x−3x2)2163(1+x-3x^{2})^{2163} is______

Find the coefficient of x2x^{2} in (x2+1x3)6(x^{2}+\dfrac{1}{x^{3}})^{6}.

If the middle term in the expansion of (x2+2)8\left(\dfrac{x}{2}+2\right)^{8} is 11201120; then the sum of possible real values of xx is

Find the middle terms in the expansion of (x2+1x)7\left(x^{2}+\dfrac {1}{x}\right)^{7}.

Find the coefficient of x3x^3 in the expansion of (2+5x)(1−2x)2(2+5x)(1-2x)^2.

The number of integral terms in the expansion of (51/2+71/8)1024{ \left( { 5 }^{ 1/2 }+{ 7 }^{ 1/8 } \right) }^{ 1024 } is................

If the second term in the expansion (a1/13+a3/2)n{ \left( { a }^{ 1/13 }+{ a }^{ 3/2 } \right) }^{ n } is 14a5/214{ a }^{ 5/2 }, then the value of nC3nC2\cfrac { { _{ }^{ n }{ C } }_{ 3 } }{ { _{ }^{ n }{ C } }_{ 2 } } is .........

If nn is odd natural number, then n∑r=0(−1)rnCr\sum _{ r=0 }^{ n }{ \cfrac { { \left( -1 \right) }^{ r } }{ { _{ }^{ n }{ C } }_{ r } } } equals................

Coefficient of x^3x^3 and x^4x^4 in { \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 } \right) }^{ 199 }{ \left( x-1 \right) }^{ 201 }{ \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 } \right) }^{ 199 }{ \left( x-1 \right) }^{ 201 } are .

2.

If the sum of the coefficients in the expansion of {(x+y)}^{n}{(x+y)}^{n} is 40964096, find the greatest coefficient in the expansion.

The greatest binomial coefficient in the expansion of \displaystyle \left ( a+b \right )^{n}\displaystyle \left ( a+b \right )^{n} is \displaystyle ^{k}C_{m}\displaystyle ^{k}C_{m} given that the sum of all the coefficients is equal to 40964096. Find k-mk-m ?

Find the negative of middle term in the expansion of\displaystyle \left ( \frac{2x}{3}-\frac{3}{2x} \right )^{6}\displaystyle \left ( \frac{2x}{3}-\frac{3}{2x} \right )^{6}

If \displaystyle ^{n+1}C_{2}+2\left ( ^2C_{2}+^3C_{2}+^4C_{2}+.....+^nC_{2} \right )=1^{2}+2^{2}+3^{2}+....+100^{2}\displaystyle ^{n+1}C_{2}+2\left ( ^2C_{2}+^3C_{2}+^4C_{2}+.....+^nC_{2} \right )=1^{2}+2^{2}+3^{2}+....+100^{2} then find sum of digits of n ?

Number of terms in the expansion of (x^{1/3} + x^{2/5})^{40}(x^{1/3} + x^{2/5})^{40} with integral power of x is equal to

Find the middle term in the expansion of (5x\, -\, 7y)^7(5x\, -\, 7y)^7.

Find the 7th term of \left (\dfrac {4x}{5}-\dfrac {5}{2x}\right )^9\left (\dfrac {4x}{5}-\dfrac {5}{2x}\right )^9.

Find the number of rational terms in the expansion of \left ( 9^{1/4}+8^{1/6} \right )^{1000}.\left ( 9^{1/4}+8^{1/6} \right )^{1000}.

\displaystyle \left ( 3a-\frac{a^{3}}{6} \right )^{9}\displaystyle \left ( 3a-\frac{a^{3}}{6} \right )^{9}

Find the middle term in the expansion of (2x + 3y)(2x + 3y)^8^8.

Find the middle term(s) in the expansion of:\left (1-\dfrac {x^2}{2}\right )^{14}\left (1-\dfrac {x^2}{2}\right )^{14}

Find the middle term(s) in the expansion of\displaystyle \left ( 1-\frac{x^{2}}{2} \right )^{14}\displaystyle \left ( 1-\frac{x^{2}}{2} \right )^{14}

Find the 13^{th}13^{th} term in the expansion of \left ( 9x - \dfrac{1}{3\sqrt x} \right )^{18}, x \neq 0\left ( 9x - \dfrac{1}{3\sqrt x} \right )^{18}, x \neq 0

Write the general term in the expansion of (x^2 - y)^6(x^2 - y)^6

Find the 4^{th}4^{th} term in the expansion of (x-2y)^{12}(x-2y)^{12}

Find the middle terms in the expansion of \left (3- \dfrac{x^3}{6} \right )^7\left (3- \dfrac{x^3}{6} \right )^7

Find the middle terms in the expansion of \left (\dfrac {x}{3}+ 9y\right)^{10}\left (\dfrac {x}{3}+ 9y\right)^{10}

Write the middle terms in the expansion of \left(\dfrac{3x}{7}-2y\right)^{10}\left(\dfrac{3x}{7}-2y\right)^{10}.

If PP and QQ are the sum of odd terms and the sum of even terms respectively, in the expansion of (x + a)^{n}(x + a)^{n} then prove that(i) P^{2} - Q^{2} = (x^{2} - a^{2})^{n}P^{2} - Q^{2} = (x^{2} - a^{2})^{n}(ii) 4PQ = (x + a)^{2n} - (x - a)^{2n}4PQ = (x + a)^{2n} - (x - a)^{2n}

Prove that C_0 +C_1 +C_2 +C_3 + ...... + 2^n . C_n = 3^nC_0 +C_1 +C_2 +C_3 + ...... + 2^n . C_n = 3^n.

Find the fifth term of { \left( a+{ 2x }^{ 3 } \right) }^{ 17 }{ \left( a+{ 2x }^{ 3 } \right) }^{ 17 }.

Write down and simplify:The 10th term of { \left( 1-2x \right) }^{ 12 }{ \left( 1-2x \right) }^{ 12 }

Find the fourteenth term of { \left( 3-a \right) }^{ 15 }{ \left( 3-a \right) }^{ 15 }.

Find the middle term of { \left( 1-\cfrac { { x }^{ 2 } }{ 2 } \right) }^{ 14 }{ \left( 1-\cfrac { { x }^{ 2 } }{ 2 } \right) }^{ 14 }.

Expand \cfrac { 7+x }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } \cfrac { 7+x }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } in ascending powers of xx and find the general term.

The 2nd, 3rd and 4th terms in the expansion of {(x+y)}^{n}{(x+y)}^{n} are 240,720, 1080240,720, 1080 respectively; find x,y,nx,y,n.

Write down and simplify:The 4th term of { { \left( \cfrac { a }{ 3 } +9b \right) }^{ 10 } }{ { \left( \cfrac { a }{ 3 } +9b \right) }^{ 10 } }.

Show that the middle term in the expansion of { \left( 1+x \right) }^{ 2n }{ \left( 1+x \right) }^{ 2n } is\cfrac { 1.3.5....(2n-1) }{ |\underline { n } } { 2 }^{ n }{ x }^{ n }\cfrac { 1.3.5....(2n-1) }{ |\underline { n } } { 2 }^{ n }{ x }^{ n }.

Find the(p+2)(p+2)th term from the end in { \left( x-\cfrac { 1 }{ x } \right) }^{ 2n+1 }{ \left( x-\cfrac { 1 }{ x } \right) }^{ 2n+1 }.

Find the rth term from the end in { \left( x+a \right) }^{ n }{ \left( x+a \right) }^{ n }.

Write down and simplify:The 25th term of { \left( 5x+8y \right) }^{ 30 }{ \left( 5x+8y \right) }^{ 30 }

Write down and simplify:The 12th term of { \left( 2x-1 \right) }^{ 13 }{ \left( 2x-1 \right) }^{ 13 }.

Write down and simplify:The 5th term of { \left( 2a-\cfrac { b }{ 3 } \right) }^{ 8 }{ \left( 2a-\cfrac { b }{ 3 } \right) }^{ 8 }.

If (1+x)^n=c_0+c_1x+c_2x^2+....+c_nx^n(1+x)^n=c_0+c_1x+c_2x^2+....+c_nx^n, find the value of 1^2c_1+2^2c_2+3^2c_3+.....+n^2c_n1^2c_1+2^2c_2+3^2c_3+.....+n^2c_n.

2.C_0+5.C_1 +8.C_2 + ...+ (3n+2)C_n = (3n+4)2^{n-1}2.C_0+5.C_1 +8.C_2 + ...+ (3n+2)C_n = (3n+4)2^{n-1}

\sum_{k=1}^{3n} 6 ^nC_{2\, k-1}(-3)^k\sum_{k=1}^{3n} 6 ^nC_{2\, k-1}(-3)^k is equal to:

Prove that ^{n}C_{r} + ^{n - 1}C_{r} + ^{n - 2}C_{r} + ..... + ^{r}C_{r} = ^{n + 1}C_{r + 1}^{n}C_{r} + ^{n - 1}C_{r} + ^{n - 2}C_{r} + ..... + ^{r}C_{r} = ^{n + 1}C_{r + 1}.

Write down and simplify 6^{th}6^{th} term in the expansion of \displaystyle \left( \frac{2x}{3} + \frac{3y}{2} \right)^9\displaystyle \left( \frac{2x}{3} + \frac{3y}{2} \right)^9.

The last term in (2^{1/3} \, + \, 2^{-1/2})^n \, is \, \left(\dfrac{1}{3(9)^{1/3}}\right)^{log_33}(2^{1/3} \, + \, 2^{-1/2})^n \, is \, \left(\dfrac{1}{3(9)^{1/3}}\right)^{log_33} then show that the fifth term is 210

Find the third term of the expansion of \displaystyle\, \left ( z^2 + \frac{1}{z} \sqrt[3]{z} \right )^n\displaystyle\, \left ( z^2 + \frac{1}{z} \sqrt[3]{z} \right )^n, if the sum of all the binomial coefficients is equal to 2048.

Find the middle term of the expansion of \displaystyle\, \left ( \sqrt{x} - \frac{1}{x} \right )^6\displaystyle\, \left ( \sqrt{x} - \frac{1}{x} \right )^6

Find the second term of the binomial expansion of \displaystyle\, \left ( \sqrt[13]{a} + \frac{a}{\sqrt{a^{-1}}} \right )^m\displaystyle\, \left ( \sqrt[13]{a} + \frac{a}{\sqrt{a^{-1}}} \right )^m, if \displaystyle\, C_{3}^{m} : C_{2}^{m} = 4 : 1\displaystyle\, C_{3}^{m} : C_{2}^{m} = 4 : 1

If the coefficient of a^{r-1}, a^ra^{r-1}, a^r and a^{r + 1}a^{r + 1} in the expansion of (1 + a)^n(1 + a)^n are in arithmetic progression, prove that n^2 - n(4r + 1) + 4r^2 - 2 = 0n^2 - n(4r + 1) + 4r^2 - 2 = 0.

In the expansion of (7^{1/3} \, + \, 11^{1/9})^{6561}(7^{1/3} \, + \, 11^{1/9})^{6561} prove that three will be only 730730 term which are free from radicals

If a_0 \, , \, a_1 \, , \, a_2a_0 \, , \, a_1 \, , \, a_2....be the coefficients in the expansion of (1 + x + x(1 + x + x^2)^n^2)^n in ascending powers xx , then prove that :a_r \, = \, a_{2n \, - \, r}a_r \, = \, a_{2n \, - \, r}

Find the middle term(s) of (\frac{x^{3/2}y}{2} + \frac{2}{xy^{3/2}})^{13}(\frac{x^{3/2}y}{2} + \frac{2}{xy^{3/2}})^{13}

Find the coefficient of x^{25}x^{25} in expansion of expression \displaystyle \sum_{r = 0}^{50} {^{50}C}_r (2x - 3)^r (2 - x)^{50 -r}\displaystyle \sum_{r = 0}^{50} {^{50}C}_r (2x - 3)^r (2 - x)^{50 -r}.

Find the middle term:(i) \left[3x - \dfrac{2}{x}\right]^{15}\left[3x - \dfrac{2}{x}\right]^{15}

Find the coefficient of a^4a^4 in the product (1 + 2a)^4 (2 - a)^5(1 + 2a)^4 (2 - a)^5 using binomial theorem.

Evaluate {(\sqrt{3}+\sqrt{2})}^{6} -{(\sqrt{3}-\sqrt{2})}^{6}.{(\sqrt{3}+\sqrt{2})}^{6} -{(\sqrt{3}-\sqrt{2})}^{6}.

In the expansion of {\left( {{x^3} - \dfrac{1}{{{x^2}}}} \right)^2}{\left( {{x^3} - \dfrac{1}{{{x^2}}}} \right)^2}, where n is a positive integer, the sum of the coefficients of {x^6}{x^6} is 1.1.

Find the middle terms in the expansion of $(5x-7y)^7$ .

Find the 28th term of $(5x+8y)^{30}$ .

Show that the expansion of $\left( x^2 + \dfrac{1}{x} \right)^{12}$ does not contain any term involving $x^{-1}$ .

Find the middle term in the expansion ${ \left( \cfrac { x }{ a } +\cfrac { y }{ b } \right) }^{ 6 }$ .

Write down and simplify:
The 4th term of ${ \left( x-5 \right) }^{ 13 }$

Find the $13^{th}$ term in the expansion of ${ \left( 9x-\cfrac { 1 }{ 3\sqrt { x } } \right) }^{18 }$ .

The sum of the coefficients in the first three terms of the expansion of $\displaystyle\, \left ( x^2 - \frac{2}{x} \right )^m$ is equal to $97$ . Find the term of the expansion containing $x^4$ .

Write general term of this:-
$2x{\left( {3 + 2{x^2}} \right)^{20}}$

Find the middle term of ${ \left( \cfrac { a }{ x } +\cfrac { x }{ a } \right) }^{ 10 }$ .

Show that $\displaystyle ^{ n }C_{ 0 } -\ ^{ n }C_{ 1 }+^{ n }C_{ 2 }\ -.....=0$

Write down and simplify:
The 7th term of ${ \left( \cfrac { 4x }{ 5 } -\cfrac { 5 }{ 2x } \right) }^{ 9 }\quad$

Find the sum $\displaystyle \sum_{r=1}^{n}\dfrac {r {^{n}C_{r}}}{{^{n}C_{r-1}}}$ .

Find the $7 ^ { th }$ term of $\left( 3 x ^ { 2 } - \frac { 1 } { 3 } \right) ^ { 10 }$ .

If $4^{th}$ term in the expansion of $\left(ax+\dfrac{1}{x}\right)^n$ is $\dfrac{5}{2}$ , then find the value of $a$ and $n$ .

Find the coefficient of $x^5$ in the expansion of the product $(1+2x)^6(1-x)^7$ .

$^{ 15 }{ C_{ 3 } }{ +^{ 15 } }{ C_{ 5 } }+.......{ +^{ 15 } }{ C_{ 15 } }$

Show that the middle term in the expansion of $(1 + x)^{n}$ is $6x^2$ if $n=4$

The sum of the rational terms in the expansion of $(\sqrt{2}+3^{1/5})^{10}$ is _______.

The sum of the coefficient of the polynomial $(1+x-3x^{2})^{2163}$ is______

Find the coefficient of $x^{2}$ in $(x^{2}+\dfrac{1}{x^{3}})^{6}$ .

If the middle term in the expansion of $\left(\dfrac{x}{2}+2\right)^{8}$ is $1120$ ; then the sum of possible real values of $x$ is

Find the middle terms in the expansion of $\left(x^{2}+\dfrac {1}{x}\right)^{7}$ .

Find the coefficient of $x^3$ in the expansion of $(2+5x)(1-2x)^2$ .

The number of integral terms in the expansion of ${ \left( { 5 }^{ 1/2 }+{ 7 }^{ 1/8 } \right) }^{ 1024 }$ is................

If the second term in the expansion ${ \left( { a }^{ 1/13 }+{ a }^{ 3/2 } \right) }^{ n }$ is $14{ a }^{ 5/2 }$ , then the value of $\cfrac { { _{ }^{ n }{ C } }_{ 3 } }{ { _{ }^{ n }{ C } }_{ 2 } }$ is .........

If $n$ is odd natural number, then $\sum _{ r=0 }^{ n }{ \cfrac { { \left( -1 \right) }^{ r } }{ { _{ }^{ n }{ C } }_{ r } } }$ equals................

Coefficient of $x^3$ and $x^4$ in ${ \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 }+{ x }^{ 4 } \right) }^{ 199 }{ \left( x-1 \right) }^{ 201 }$ are .

If the sum of the coefficients in the expansion of ${(x+y)}^{n}$ is $4096$ , find the greatest coefficient in the expansion.

The greatest binomial coefficient in the expansion of $\displaystyle \left ( a+b \right )^{n}$ is $\displaystyle ^{k}C_{m}$ given that the sum of all the coefficients is equal to $4096$ . Find $k-m$ ?

Find the negative of middle term in the expansion of
$\displaystyle \left ( \frac{2x}{3}-\frac{3}{2x} \right )^{6}$

If $\displaystyle ^{n+1}C_{2}+2\left ( ^2C_{2}+^3C_{2}+^4C_{2}+.....+^nC_{2} \right )=1^{2}+2^{2}+3^{2}+....+100^{2}$ then find sum of digits of n ?

Number of terms in the expansion of $(x^{1/3} + x^{2/5})^{40}$ with integral power of x is equal to

Find the middle term in the expansion of $(5x\, -\, 7y)^7$ .

Find the 7th term of $\left (\dfrac {4x}{5}-\dfrac {5}{2x}\right )^9$ .

Find the number of rational terms in the expansion of $\left ( 9^{1/4}+8^{1/6} \right )^{1000}.$

$\displaystyle \left ( 3a-\frac{a^{3}}{6} \right )^{9}$

Find the middle term in the expansion of $(2x + 3y)$ $^8$ .

Find the middle term(s) in the expansion of:
$\left (1-\dfrac {x^2}{2}\right )^{14}$

Find the middle term(s) in the expansion of
$\displaystyle \left ( 1-\frac{x^{2}}{2} \right )^{14}$

Find the $13^{th}$ term in the expansion of $\left ( 9x - \dfrac{1}{3\sqrt x} \right )^{18}, x \neq 0$

Write the general term in the expansion of $(x^2 - y)^6$

Find the $4^{th}$ term in the expansion of $(x-2y)^{12}$

Find the middle terms in the expansion of $\left (3- \dfrac{x^3}{6} \right )^7$

Find the middle terms in the expansion of $\left (\dfrac {x}{3}+ 9y\right)^{10}$

Write the middle terms in the expansion of $\left(\dfrac{3x}{7}-2y\right)^{10}$ .

If $P$ and $Q$ are the sum of odd terms and the sum of even terms respectively, in the expansion of $(x + a)^{n}$ then prove that
(i) $P^{2} - Q^{2} = (x^{2} - a^{2})^{n}$
(ii) $4PQ = (x + a)^{2n} - (x - a)^{2n}$

Prove that $C_0 +C_1 +C_2 +C_3 + ...... + 2^n . C_n = 3^n$ .

Find the fifth term of ${ \left( a+{ 2x }^{ 3 } \right) }^{ 17 }$ .

Write down and simplify:
The 10th term of ${ \left( 1-2x \right) }^{ 12 }$

Find the fourteenth term of ${ \left( 3-a \right) }^{ 15 }$ .

Find the middle term of ${ \left( 1-\cfrac { { x }^{ 2 } }{ 2 } \right) }^{ 14 }$ .

Expand $\cfrac { 7+x }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) }$ in ascending powers of $x$ and find the general term.

The 2nd, 3rd and 4th terms in the expansion of ${(x+y)}^{n}$ are $240,720, 1080$ respectively; find $x,y,n$ .

Write down and simplify:
The 4th term of ${ { \left( \cfrac { a }{ 3 } +9b \right) }^{ 10 } }$ .

Show that the middle term in the expansion of ${ \left( 1+x \right) }^{ 2n }$ is
$\cfrac { 1.3.5....(2n-1) }{ |\underline { n } } { 2 }^{ n }{ x }^{ n }$ .

Find the $(p+2)$ th term from the end in ${ \left( x-\cfrac { 1 }{ x } \right) }^{ 2n+1 }$ .

Find the rth term from the end in ${ \left( x+a \right) }^{ n }$ .

Write down and simplify:
The 25th term of ${ \left( 5x+8y \right) }^{ 30 }$

Write down and simplify:
The 12th term of ${ \left( 2x-1 \right) }^{ 13 }$ .

Write down and simplify:
The 5th term of ${ \left( 2a-\cfrac { b }{ 3 } \right) }^{ 8 }$ .

If $(1+x)^n=c_0+c_1x+c_2x^2+....+c_nx^n$ , find the value of $1^2c_1+2^2c_2+3^2c_3+.....+n^2c_n$ .

$2.C_0+5.C_1 +8.C_2 + ...+ (3n+2)C_n = (3n+4)2^{n-1}$

$\sum_{k=1}^{3n} 6 ^nC_{2\, k-1}(-3)^k$ is equal to:

Prove that $^{n}C_{r} + ^{n - 1}C_{r} + ^{n - 2}C_{r} + ..... + ^{r}C_{r} = ^{n + 1}C_{r + 1}$ .

Write down and simplify $6^{th}$ term in the expansion of $\displaystyle \left( \frac{2x}{3} + \frac{3y}{2} \right)^9$ .

The last term in $(2^{1/3} \, + \, 2^{-1/2})^n \, is \, \left(\dfrac{1}{3(9)^{1/3}}\right)^{log_33}$ then show that the fifth term is 210

Find the third term of the expansion of $\displaystyle\, \left ( z^2 + \frac{1}{z} \sqrt[3]{z} \right )^n$ , if the sum of all the binomial coefficients is equal to 2048.

Find the middle term of the expansion of $\displaystyle\, \left ( \sqrt{x} - \frac{1}{x} \right )^6$

Find the second term of the binomial expansion of $\displaystyle\, \left ( \sqrt[13]{a} + \frac{a}{\sqrt{a^{-1}}} \right )^m$ , if $\displaystyle\, C_{3}^{m} : C_{2}^{m} = 4 : 1$

If the coefficient of $a^{r-1}, a^r$ and $a^{r + 1}$ in the expansion of $(1 + a)^n$ are in arithmetic progression, prove that $n^2 - n(4r + 1) + 4r^2 - 2 = 0$ .

In the expansion of $(7^{1/3} \, + \, 11^{1/9})^{6561}$ prove that three will be only $730$ term which are free from radicals

If $a_0 \, , \, a_1 \, , \, a_2$ ....be the coefficients in the expansion of $(1 + x + x$ $^2)^n$ in ascending powers $x$ , then prove that :
$a_r \, = \, a_{2n \, - \, r}$

Find the middle term(s) of $(\frac{x^{3/2}y}{2} + \frac{2}{xy^{3/2}})^{13}$

Find the coefficient of $x^{25}$ in expansion of expression $\displaystyle \sum_{r = 0}^{50} {^{50}C}_r (2x - 3)^r (2 - x)^{50 -r}$ .

Find the middle term:
(i) $\left[3x - \dfrac{2}{x}\right]^{15}$

Find the coefficient of $a^4$ in the product $(1 + 2a)^4 (2 - a)^5$ using binomial theorem.

Evaluate ${(\sqrt{3}+\sqrt{2})}^{6} -{(\sqrt{3}-\sqrt{2})}^{6}.$

In the expansion of ${\left( {{x^3} - \dfrac{1}{{{x^2}}}} \right)^2}$ , where n is a positive integer, the sum of the coefficients of ${x^6}$ is $1.$

Find the sum $\sum _{ r=1 }^{ n }{ \overset { n+r }{ \underset { \quad \quad r }{ C }}}$ .

Find the sum of $\sum _{ 0\le i\le j\le n } \sum { j\overset { n }{ \underset { \quad i }{ C } } }$