Class 11 Commerce Maths - Binomial Theorem

Find the sum of all rational terms in the expansion of ${ \left( { 3 }^{ 1/5 }+{ 2 }^{ 1/3 } \right) }^{ 15 }$ .

$\displaystyle T_{r+1}=\:^{15}C_{r}3^{3-\dfrac{r}{5}}2^{\dfrac{r}{3}}$
Hence only

$T_{1}$ and

$T_{16}$ will be the rational terms.

$T_{0+1}=\:^{15}C_{0}3^{3}$

$=27$

$...(a)$

$T_{15+1}=\:^{15}C_{15}2^{5}$

$=32$

$...(b)$
Hence,

$a+b$ implies

$27+32$

$=59$

Find the number of terms in the expansion of ${ \left( \sqrt [ 4 ]{ 9 } +\sqrt [ 6 ]{ 8 } \right) }^{ 500 }$ which are integers.

$\therefore$ General term

${ _{ }^{ }{ T } }_{ r+1 }$

$={ _{ }^{ 500 }{ C } }_{ r }{ \left( { 3 }^{ 1/2 } \right) }^{ 500-r }{ \left( { 2 }^{ 1/2 } \right) }^{ r }$

$={ _{ }^{ 500 }{ C } }_{ r }{ 3 }^{ 250-r/2 }.{ 2 }^{ r/2 }$

For

$r=0,2,4,6,8,10,12,.....500,$

Powers of

$3$ and

$2$ are integers.
Hence, the integers terms

$=251$

Alternative:

$(9^{1/4}+8^{1/6})^{500}$

$=(3^{1/2}+2^{1/2})^{500}$
Now, the total number of integral terms will be

$1+\dfrac{500}{L.C.M(2,2)}$

$=1+\dfrac{500}{2}$

$=1+250$

$=251$

Find the coefficient of $x^5$ in $(x+3)^8$

It is known that

$(r+1)$ th term

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

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

$T_{r+1} = ^nC_r a^{a-r}b^r$
Assuming that

$x^5$ occurs in the

$(r+1)$ th term of expansion

$(x+3)^8$ we obtain

$T_{r+1} = ^8C_r(x) ^{8-r} (3)^r$
Comparing the indices of

$x$ in

$x^5$ and in

$T_{r+1}$ , we obtain

$r= 3$
Thus, the coefficient of

$x^5$ is

$\displaystyle ^8C_3(3)^3 = \frac{8!}{3!5!}x 3^3 = \frac{8.7.6.5!}{3.2.5!}.3^3 = 1512$

The coefficient of three consecutive terms in the expansion of ${ \left( 1+a \right) }^{ n }$ are in ratio $1:7:21$ , then find the value of $n$ .

First term =

$^nC_r$

Second term =

$^nC_{r+1}$

Third term =

$^nC_{r+2}$

$\dfrac{1}{7}=\dfrac{^nC_r}{^nC_{r+1}}=\dfrac{n!}{r!|n-r|!}=\dfrac{|r+1|(n-r-1)1}{n!}($

$\dfrac{1}{7}=\dfrac{r+1}{n-r}$

$n-r=7r+7$ ......(1)

$\dfrac{7}{21}=\dfrac{1}{3}=\dfrac{n!}{(r+1)!(n-r-1)!}\times \dfrac{(r+2)!(n-r-2)}{n!}$

$\dfrac{1}{3}=\dfrac{r+2}{n-r-1}$

$n-r-1=3r+6$

$n=4r+7$ ...........(2)

$n=8r+7$ ........(1)

$r=0$

$\therefore n=7$

$\dfrac { { C }_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } +\dfrac { { C }_{ 5 } }{ 6 } +...=\dfrac { { 2 }^{ n }-1 }{ n+1 }$

The

$r_{th}$ term in the given expansion is

$T_{r}=\dfrac{\displaystyle^{n}C_{2r-1}}{2r}$

Since

$\dfrac{1}{r+1}\cdot \displaystyle^nC_{r}=\dfrac{1}{n+1}\cdot \displaystyle^{n+1}C_{r+1}$

$\therefore T_{r}=\dfrac{\displaystyle^n C_{2r-1}}{2r}=\dfrac{1}{n+1}\cdot \displaystyle^{n+1}C_{2r}$

$\therefore \dfrac{\displaystyle^nC_{1}}{2}+\dfrac{\displaystyle^nC_{3}}{4}+\dfrac{\displaystyle^nC_{5}}{6}+......=\dfrac{1}{n+1}[\displaystyle^{n+1}C_{2}+\displaystyle^{n+1}C_{4}+.....]$

$=\dfrac{1}{n+1}[2^{n+1-1}-\displaystyle^nC_{0}]=\dfrac{2^n-1}{n+1}$

Find the term of the expansion of $(a + b)^{50}$ which is the greatest in absolute value if $\displaystyle\, \left | a \right | = \sqrt{3}\left | b \right |$

Given

$(a+b)^{50}$

General term of given expansion is

$T_{r+1}=\sum_{r=0}^{50}\displaystyle^{50}C_{r}(a)^{50-r}(b)^r$

Here

$a=\sqrt{3}b$

$T_{r+1}=\sum_{r=0}^{50}\displaystyle^{50}C_{r}(\sqrt{3}b)^{50-r}(b)^r$

$T_{r+1}=\sum_{r=0}^{50}\displaystyle^{50}C_{r}(\sqrt{3})^{50-r}(b)^{r+50-r}$

$T_{r+1}=\sum_{r=0}^{50}\displaystyle^{50}C_{r}(\sqrt{3})^{50-r}(b)^{50}$

Middle term

$25^{th}$ has maximum value of binomial coefficient

$T_{25}=\displaystyle^{50}C_{24}(\sqrt{3})^{26}(b)^{50}$

But we should compare it with

$26^{th}$ term

$T_{26}=\displaystyle^{50}C_{25}(\sqrt{3})^{25}(b)^{50}$

Hence comaparing

$T_{25},T_{26}$

we get

$T_{25}$ is greatest term

Prove that the following equations hold for any natural n.
Prove that $\displaystyle (C_{n}^{o})^2 \, + \, (C_{n}^{1})^2 \, + \, ... \, + \, (C_{n}^{n})^2 \, = \, C_{2n}^{n}.$

For deriving our expression

we have to first simplyfy this

$\quad \quad ({ ^nC }_{ 0 })^{ 2 }+9^n{ C }_{ 1 })^{ 2 }+(^n{ C }_{ 3 })^{ 2 }+(^n{ C }_{ 4 })^{ 2 }+.......+({ ^nC }_{ n })^{ 2 }\\ =({ C }_{ 0 }{ C }_{ n }+{ C }_{ 2 }{ C }_{ n-1 }+.........+{ C }_{ n }{ C }_{ 0 })(\therefore\ ^nC_0=^nC_n)$

$=Cofficient\quad of\quad { x }^{ n }\quad in\quad { (1+x) }^{ n }{ (1+x) }^{ n }\\ =Cofficient\quad of\quad { x }^{ n }\quad in\quad { (1+x) }^{ 2n }\\ =^{ 2n }C_{ n }$

Find the power $n$ of the binomial $\displaystyle\, \left ( \frac{x}{5} + \frac{2}{5} \right )^n$ , if the ninth term of the expansion has the greatest coefficient.

Given:

$\left ( \dfrac{x}{5} + \dfrac{2}{5} \right )^n$

General term of given expansion is

$T_{r+1}=\sum_{r=0}^{n}\displaystyle^{n}C_{r}\left ( \dfrac{x}{5} \right )^{n-r}\left ( \dfrac{2}{5} \right )^r$

$9^{th }$ term When

$r=8$

$T_{8+1}=\displaystyle^{n}C_{8}\left ( \dfrac{x}{5} \right )^{n-8}\left ( \dfrac{2}{5} \right )^8$

$T_{9}=\displaystyle^{n}C_{8}x^{n-8}2^8\left( \dfrac{1}{5} \right )^{8+n-8}$

$T_{9}=\displaystyle^{n}C_{8} x^{n-8}2^8\left( \dfrac{1}{5^n} \right )$

As we know that in the binomial expansion the middle term has the greatest value

So if n is even ,then

$\dfrac{n}{2}=9\Rightarrow n=18$

Hence

$T_{9}=\displaystyle^{18}C_{8} x^{10}2^8\left( \dfrac{1}{5^{18}} \right )$

Coefficient of above term is

$\displaystyle^{18}C_{8}2^8\left( \dfrac{1}{5^{18}} \right )$ .....

$(1)$

When n is odd then

$9^{th}$ term can be middle term of

$\dfrac{n+1}{2}=9$ and

$\dfrac{n+3}{2}=9$

$9^{th}$ term can be middle term of

$n=17$ and

$n=15$

When

$n=17$

$T_{9}=\displaystyle^{17}C_{8} x^{9}2^8\left( \dfrac{1}{5^{17}} \right )$

Coefficient

$\displaystyle^{17}C_{8} 2^8\left( \dfrac{1}{5^{17}} \right )$ ......

$(2)$

When

$n=15$

$T_{9}=\displaystyle^{15}C_{8} x^{7}2^8\left( \dfrac{1}{5^{15}} \right )$

Coefficient

$\displaystyle^{15}C_{8} 2^8\left( \dfrac{1}{5^{15}} \right )$ .....

$(3)$

Comparing equations

$(2)$ and

$(3),$ we get

$\displaystyle^{15}C_{8} 2^8\left( \dfrac{1}{5^{15}} \right )$ is greatest

But we should also compare eq (3) and (1) ,we get

$\displaystyle^{15}C_{8} 2^8\left( \dfrac{1}{5^{15}} \right )$ is greatest

Hence

$n=15$

Evaluate :
$\left( {\dfrac{{{C_0} + {C_1}}}{{{C_0}}}} \right)\left( {\dfrac{{{C_1} + {C_2}}}{{{C_1}}}} \right)\left( {\dfrac{{{C_2} + {C_3}}}{{{C_2}}}} \right)\left( {\dfrac{{{C_3} + {C_4}}}{{{C_3}}}} \right)........\left( {\dfrac{{{C_{n - 1}} + {C_n}}}{{{C_{n - 1}}}}} \right)$

We know, for any $k < n$ , ${{{C_{k + 1}}} \over {{C_k}}}$

$= {{n!\,\,\left( {n - k} \right)!k!} \over {\left( {k + 1} \right)!\left( {n - k - 1} \right)!n!}}$

$= {{\left( {n - k} \right)\left( {n - k - 1} \right)!k!} \over {\left( {k + 1} \right)k!\left( {n - k - 1} \right)!}} = {{n - k} \over {k + 1}}$

Now, ${{{C_k} + {C_{k + 1}}} \over {{C_k}}} = 1 + {{{C_{k + 1}}} \over {{C_k}}} = 1 + {{n - k} \over {k + 1}} = {{n - k + k + 1} \over {k + 1}} = {{n + 1} \over {k + 1}}$

Then, $\left( {{{{C_0} + {C_1}} \over {{C_0}}}} \right)\left( {{{{C_1} + {C_2}} \over {{C_1}}}} \right).....\left( {{{{C_n} + {C_n}} \over {{C_{n - 1}}}}} \right)$

$= {{n + 1} \over {0 + 1}}.{{n + 1} \over {1 + 1}}.....{{n + 1} \over {\left( {n - 1} \right) + 1}} = {{{{\left( {n + 1} \right)}^n}} \over {1.2......n}} = {{{{\left( {n + 1} \right)}^n}} \over {n!}}$

$\left( {{{{C_0} + {C_1}} \over {{C_0}}}} \right)\left( {{{{C_1} + {C_2}} \over {{C_1}}}} \right)....\left( {{{{C_{n - 1}} + {C_n}} \over {{C_{n - 1}}}}} \right) = {{{{\left( {n + 1} \right)}^n}} \over {n!}}$

If in the expansion of $(1+x)^{20}$ , the coefficients of $r^{th}$ and $(r+4)^{th}$ terms are equal, then r is equal to?

We are given In the expansion of

$(1+x)^{20}$ , the coefficients of

$r^{th}$ and

$(r+4)^{th}$ terms are equal.

here the any

$r^{th}$ term will be

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

now, here

$a=1$ &

$b=x$

so, the coefficient of

$r^{th}$ term is

$=\left (\dfrac {n}{r-1}\right) (1)^{n-r+1}, \boxed {n=20}$

$=\left (\dfrac {20}{r-1}\right)$

& the co-efficient of

$(r+4)^{th}$ term is

$=\left (\dfrac {n}{r+4-1}\right) (1)^{n-r-4+1}, \boxed {n=20}$

$=\left (\dfrac {20}{r+3}\right)$

so,

$\left (\dfrac {20}{r-1}\right)=\left (\dfrac {20}{r+3}\right)$

now, it

$\left (\dfrac {n}{r_1}\right)=\left (\dfrac {n}{r_2}\right)\Rightarrow \ n=r_1+r_2$

so,

$20=r-1+r+3$

$2r+2=20$

$2r=18$

$r=9$

The

$r$ is equal to

$9$

If $n> 2$ then prove that ${C}_{1}(a-1)-{C}_{2}\times(a-2)+.....+{(-1)}^{n-1}{C}_{n}(a-n)=a$ ,where ${C}_{r}={ _{ }^{ n }{ C } }_{ r }$

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

Differentiating on both sides

$-n(1-x)^{n-1}=0-C_{1}+2{C_{2}}(x)-3{C_{3}}(x)^{2}+......+n(-1)^{n}(x)^{n-1}......(2)$

When

$x=1$

$(1)$ becomes

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

$\implies C_{1}-C_{2}+C_{3}-.....+(-1)^{n-1}C_{n}=C_{0}=1$

$(2)$ becomes

$0=-C_{1}+{2}C_{2}-{3}C_{3}+....+(-1)^{n}{n}C_{n}$

$LHS=C_{1}(a-1)-C_{2}(a-2)+.......+(-1)^{n-1}C_{n}(a-n)$

$=a(C_{1}-C_{2}+C_{3}-....+(-1)^{n-1}C_{n})+(-C_{1}+2{C_2}-3{C_{3}}+....+(-1)^{n}{n}C_{n})=a(1)+0=a=RHS$

$LHS=RHS$

Hence Proved

Coefficient of $x^{10}$ in the expansion of $\left(2x^2-\dfrac{3}{x}\right)^{11}, x\neq 0$ .

1084468_1113413_ans_1d90de945eb945859ff7715432763fb0.jpg

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

1407614_1138925_ans_be1494be73d14bf4abbde1c928c5d1da.jpg

Find a positivevalue of m for which the confficient of $x^{2}$ in the expansion $\left ( 1+x \right )^{m}$ is $6$

Consider the problem

It is know that

${\left( {r + 1} \right)^{th}}$ term,

$\left( {{T_{r + 1}}} \right)$ , in the binomial expansion of

${\left( {a + b} \right)^n}$ is given by

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

Assuming that

${x^2}$ occurs in the

${\left( {r + 1} \right)^{th}}$ term of expansion

${\left( {1 + x} \right)^m}$ we obtain

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

Now comparing the indices of

$x$ in

${x^2}$ and in

${T_{r + 1}}$ we obtain

$r=2$

Therefore, the coefficient of

${x^2}$ is

$^m{C_2}$

It is given that the coefficient of

${x^2}$ in the expansion

${\left( {1 + x} \right)^m}$ is

$6$

$\begin{array}{l} \therefore \, { \, ^{ m } }{ C_{ 2 } }=6 \\ \Rightarrow \dfrac { { m! } }{ { 2!\left( { m-2 } \right) ! } } =6 \\ \Rightarrow \dfrac { { m\left( { m-1 } \right) \left( { m-2 } \right) ! } }{ { 2\times \left( { m-2 } \right) ! } } =6 \\ \Rightarrow m\left( { m-1 } \right) =12 \\ \Rightarrow { m^{ 2 } }-m-12=0 \\ \Rightarrow { m^{ 2 } }-4m+3m-12=0 \\ \Rightarrow m\left( { m-4 } \right) +3\left( { m-4 } \right) =0 \\ \Rightarrow \left( { m-4 } \right) \left( { m-3 } \right) =0 \\ \Rightarrow \left( { m-4 } \right) =0\, or\left( { m+3 } \right) =0 \\ \Rightarrow m=4\, or\, m=-3 \end{array}$

Thus, the positive value of

$m$ for which the coefficient of

${x^2}$ in the expansion

${\left( {1 + x} \right)^m}$ is

$6$ is

$4$ .

The coefficient of ${x}^{20}$ in the expansion of ${(1+{x}^{2})}^{40}({x}^{2}+2+\cfrac{1}{{x}^{2}})$ is

$x^{2}(1+x^{2})^{40}+2(1+x^{2})^{40}+\frac{1}{x^{2}}(1+x^{2})^{40}$

writing general tence for all three,

$x^{2}\times ^{40}C_{r_{1}}(x^{2})^{r_{1}},2\times ^{40}C_{2}\times (x^{2})^{r_{2}},\frac{1}{^{2}}\times ^{40}C_{3}(x^{2})^{r_{3}}$

For

$x ^{20},$

$r_{1} = 9,r_{2} = 10, r_{3} = 11$

$\Rightarrow$ cofficients

$= ^{40}C_{9}+2\times ^{40}C_{10}+^{40}C_{11}$

$^{n}C_{r}+^{n}C_{r-1} = ^{n+1}C_{r}$

$\Rightarrow$ coeff.

$= ^{41}C_{10}+^{41}C_{11} = ^{42}C_{11}$

$\Rightarrow$ No option

1124067_1204045_ans_9b861085102f454da1e12ba58810b0dd.jpg

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

The general term,

$T_{r+1}=\,^nC_rx^{n-r}y^r$

$\Rightarrow$

$T_2=\,^nC_1.x^{n-1}.y=240$ ---- ( 1 )

$\Rightarrow$

$T_3=\,^nC_2.x^{n-2}.y^2=720$

$\Rightarrow$

$T_4=\,^nC_3.x^{n-3}.y^3=1080$

$\Rightarrow$

$\dfrac{T_3}{T_2}=\dfrac{(n-1)}{2}\times \dfrac{y}{x}=3$ ---- ( 2 )

$\Rightarrow$

$\dfrac{T_4}{T_2}=\dfrac{(n-1)(n-2)}{3\times 2}\times\dfrac{x^2}{y^2}=\dfrac{9}{2}$ ---- ( 3 )

$\Rightarrow$

$\dfrac{T_4}{T_3}=\dfrac{(n-2)}{3}\times\dfrac{y}{x}=\dfrac{3}{2}$ ---- ( 4 )

Dividing equation ( 2 ) by ( 4 ) we get,

$\Rightarrow$

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

$\Rightarrow$

$3n-3=4n-8$

$\Rightarrow$

$5=n$

Substituting value of

$n$ in ( 4 ) we get,

$\Rightarrow$

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

Putting

$y=\dfrac{3}{2}x$ and

$n=5$ in equation ( 1 ) we get,

$\Rightarrow$

$5x^5\times\dfrac{3}{2}=240$

$\Rightarrow$

$x^5=32$

$\Rightarrow$

$x=2$

Now, substituting value of

$x$ in

$y=\dfrac{3}{2}x$ we get,

$\Rightarrow$

$y=3$

$\therefore$

$x=,2,\,y=3,\,n=5$

In the expansion of $(1+x)^{43}$ the coefficients of the $(2r+1)^{th}$ and the $(r+2)^{th}$ terms are equal; find $r$ .

In the expansion of

$(1+x)^{43}$

$T_{2r+1}=\,^{43}C_{2r}1^{43-2r}x^{2r}=\,^{43}C_{2r}x^{2r}$

$T_{r+2}=\,^{43}C_{r+1}1^{43-(r+1)}x^{r+1}=\,^{43}C_{r+1}x^{r+1}$

According to the question,

$\Rightarrow$

$^{43}C_{2r}=\,^{43}C_{r+1}$

So, either

$2r=r+1$ or

$2r+(r+1)=43$

$\Rightarrow$

$r=1$ or

$3r=42$

$\therefore$

$r=1$ or

$r=14$

Find the coefficient of $x^{5}$ in $\left( x + 3 \right)^{n}$

It is known that

$(r + 1)^{th}$ term,

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

$(a+b)6n$ is given by

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

Assuming that

$x^5$ occurs in the

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

$(x+3)^8$ , we obtain

$T_{r+2} = \ ^nC_r (x)^{8-r}(3)^r$

Comparing the indices of

$x$ in

$x^5$ in

$T_{r+1}$ .

we obtain

$r=3$

thus, the coefficient of

$x^5$ is

$^8C_3 (3)^3 = \dfrac{8!}{3!5!} 3^3 = \dfrac{8.7.6.5!}{3.25!} . 3^3 = 1512$ .

Find the coefficient of $a^{5} b^{7}$ in $\left (a - 2b \right )^{12}$

It is known that

$\left (r + 1 \right )^{th}$ term,

$\left ( T_{r-2} \right )$ , in the binomial expansion of

$\left (a + b \right )^{n}$ is given by

$T_{r+1} = {^n}C_r a^{n-r} b^{r}$
Assuming that

$a^{5} b^{7}$ occurs in the

$\left ( r + 1 \right )^{12}$ term of the expansion

$\left ( a - 2b \right )^{12}$ , we obtain

$T_{r+1} = {^12}C_2 (a)^{12-r} (-2b)^{r} = {^12}C_2 (-2)^{r} (a)^{12-r} (b)^{r}$
Comparing the indices of a and b in

$a^{5} b^{7}$ in

$T_{2*}$
We obtain r = 7
Thus, the coefficient of

$a^{5} b^{7}$ is

${^12}C_r (-2)^{7} = \frac{12!} {7!5!} \cdot 2^{7} = \frac{12 \cdot 11.10 \cdot 9 \cdot 8 \cdot 7!} {5 \cdot 4.3 \cdot 2 \cdot 7!} \cdot(-2)^{7} = - (792) (128) = -101376$

Find the number of term in the expansion of $(x + a)^{200}+ (x - a)^{200}$ after simplification.

$\because (x + a)^n + (x – a)^n$

$= 2[\,^nC_0 x^na^0 + \,^nC_0 x^{n-2} - a^2 + \,^nC_4x^{n-4} .a^4 +…]$
where n is an even number then
[Number of terms in the expansion of

$(x + a)^n + (x – a)^n$ ]
will be

$\bigg(\dfrac{n}{2}+1\bigg)$
Here,

$n = 200$
Hence, number of terms in the expansion
Will be

$=\bigg(\dfrac{200}{2}+1\bigg)=101$

Hence, the number of terms is

$101$ in the expansion.

Find the coefficient of $x^{256}$ in $(1 - x)^{101} (x^{2} + x + 1)^{100}$

$(1 - x)^{101} (x^{2} + x + 1)^{100}$

$= (1 - x)(1 - x)^{100} (x^{2} + x + 1)^{100}$

$= (1 - x)(1 - x^{3})^{100}$

$= (1 - x^{3})^{100} - x(1 - x^{3})^{100}$
First term will not have

$x^{256}$ .
Coefficient of

$x^{256}$ is

$-1\times$ coefficient of

$x^{255}$ in

$(1 - x^{3})^{100}$

$[\because\,x^{255}=(x^3)^{85}]$

$= -1(-^{100}C_{85}) = ^{100}C_{85}$

If $a_n$ denotes the coefficient of $x^n$ in $P(x)=(1+x+2x^2+....+25x^{25})^2$ , find $\dfrac {a_5}{5}$ .

$\displaystyle P(x)=(1+x+2x^2+3x^3+....+25x^{25})(1+x+2x^2+3x^3+....+25x^{25})$
Looking at the terms that can have

$x^5$ , we get

$\displaystyle a_5=\text{coeff of}\: x^5=1.5+1.4+2.3+3.2+4.1+5=30$

$\Rightarrow \dfrac {a_5}{5}=6$

If the sixth term in the expansion of ${ \left( \cfrac { 1 }{ { x }^{ 8/3 } } +{ x }^{ 2 }\log _{ 10 }{ x } \right) }^{ 8 }$ is $5600$ , find $x.$

Step 1:Apply formula of general term

Given :

$(\dfrac { 1 }{ { x }^{ 8/3 } } +{ x }^{ 2 }\log { x)^{ 8 } }$

we know that general term of expansion of

$(a+b)^n$ is

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

$T_{ r+1 }=^nC_{ r } (\dfrac { 1 }{ { x }^{ 8/3 } } )^{ 8-r }(x^{ 2 }\log { x)^{ r } }$

Step2: Simplify to evaluate required unknown

$T_{ 6}=$

$\quad ^{ 8 }C_{ 5 }[\dfrac { 1 }{ { x }^{ \tfrac { 8 }{ 3 } } } ]^{ 3 }({ x }^{ 2 }\log { x)^{ 5 } } =5600$

$\Rightarrow { x }^{ -8+10 }{ (\log { x } ) }^{ 5 }\times 8\times 7=5600$

$\Rightarrow { x }^{ 2 }(\log { x } )^{ 5 }=100$

$\Rightarrow x=10$

Hence value of x is 10

The value of $p$ , for which coefficient of $x^{50}$ in the expression $(1 + x)^{1000} + 2x (1 + x)^{999} + 3x^{2} (1 + x)^{998} + .... + 1001 x^{1000}$ is equal to $^{1002}C_{p}$ , is

Let

$S=\sum^{1000}_{r=0}{(1+r)x^r(1+x)^{1000-r}}$

So,

$S-\dfrac x{(1+x)}S=(\sum^{1000}_{r=0}{x^r(1+x)^{1000-r})-1001\dfrac {x^{1001}}{(1+x)}}=(\dfrac {(1+x)^{1000}(1-(\dfrac x{1+x})^{1001})}{1-\dfrac x{1+x}})-1001\dfrac {x^{1001}}{1+x}$

$\Rightarrow S(1-\dfrac x{1+x})=((1+x)^{1001}-x^{1001})-1001\dfrac {x^{1001}}{1+x}$

$\Rightarrow S=(1+x)((1+x)^{1001}-x^{1001})-1001x^{1001}=(1+x)^{1002}-1002x^{1001}-x^{1002}$

So,

$S=(\sum^{1002}_{r=0}{^{1002}C_rx^r})-1002x^{1001}-x^{1002}$

So, coefficient of

$x^{50}$ in the expansion will be,

$^{1002}C_{50}$

So,

$p=50$ or

$p=1002-50=952$ (

$\because ^nC_r=^nC_{n-r}$ ).

Find $n$ , if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${ \left( \sqrt [ 4 ]{ 2 } +\cfrac { 1 }{ \sqrt [ 4 ]{ 3 } } \right) }^{ n }$ is $\sqrt { 6 } :1$

$T_5$ in

$\left[\sqrt[4]{2}+\dfrac{1}{\sqrt[4]{3}}\right]^n=nC_4(\sqrt[4]{2})^{n-4}\cdot \left(\dfrac{1}{\sqrt[4]{3}}\right)^4=nC_42^{\dfrac{n-4}{4}}\cdot \dfrac{1}{3}\quad ...(1)$

Total number of terms

$=n+1$

Fifth term from the end

$=[(n+1)-5+1]^{th}$ term from the beginning.

$\Rightarrow (n-3)^{th}$ term from the beginning.

$\Rightarrow T_{n-3}=nC_{n-4}\left(\sqrt[4]{2}\right)^{n-(n-4)}\left(\dfrac{1}{\sqrt[4]{3}}\right)^{n-4}=nC_{n-4}\cdot 2\left(\dfrac{1}{3}\right)^{\dfrac{n-4}{4}}\quad ...(2)$

Given

$\dfrac{T_5}{T_{n-3}}=\dfrac{\sqrt{6}}{1}$

$\dfrac{nC_42^{\dfrac{n-4}{4}}\cdot \dfrac{1}{3}}{nC_{n-4}\cdot 2\left(\dfrac{1}{3}\right)^{\dfrac{n-4}{4}}}=\dfrac{\sqrt{6}}{1}$

$\Rightarrow \dfrac{nC_4}{nC_{n-4}}(2)^{\dfrac{n-4}{4}-1}\cdot \left(\dfrac{1}{3}\right)^{1-\dfrac{n-4}{4}}=\sqrt{6}$

$\Rightarrow \dfrac{nC_4}{nC_{n-4}}(2)^{\dfrac{n-8}{4}}\cdot \left(\dfrac{1}{3}\right)^{\dfrac{8-n}{4}}=\sqrt{6}$

$\Rightarrow \dfrac{nC_4}{nC_{n-4}}(2)^{\dfrac{n-8}{4}}\cdot (3)^{-\dfrac{8-n}{4}}=\sqrt{6}$

$\Rightarrow \dfrac{nC_4}{nC_{n-4}}(2)^{\dfrac{n-8}{4}}\cdot (3)^{\dfrac{n-8}{4}}=\sqrt{6}$

$\Rightarrow \dfrac{nC_4}{nC_{n-4}}(6)^{\dfrac{n-8}{4}}=\sqrt{6}$

Comparing powers of

$6$

$\Rightarrow \dfrac{n-8}{4}=\dfrac{1}{2}$

$\Rightarrow n-8=2$

$\therefore n=10$

Let $X = (^{10}C_{1})^{2} + (^{10}C_{2})^{2} + 3(^{10}C_{3})^{2} + .... + 10 (^{10}C_{10})^{2}$ , where $^{10}C_{r}, r\epsilon \left \{1, 2, ..., 10\right \}$ denote binomial coefficients. Then, the value of $\dfrac {1}{1430}X$ is

$X = \displaystyle \sum_{r = 0}^{n} r. (^{n}C_{r})^{2}; n = 10$

$\Rightarrow X = n. \displaystyle \sum_{r = 0}^{n} \ ^{n}C_{r} . ^{n - 1}C_{r - 1}$

$\Rightarrow X = n. \displaystyle \sum_{r = 0}^{n}\ ^{n}C_{n - r}. ^{n - 1}C_{r - 1}$

$\Rightarrow X = n . ^{2n - 1}C_{n - 1}; n = 10$

$\Rightarrow X = 10 . ^{19}C_{9}$

$\Rightarrow \dfrac {X}{1430} = \dfrac {1}{143} . ^{19}C_{9}$

$= 646$ .

Evaluate $\sum_{r = 0 }^{n } \, (r \, + \, 1)^3 \, C_r \, whee C_r \, = \, ^nC_r$

$\sum_{r=0}^{n} \, (C_r) \, = \, C_0 \, + \, 2^3c_1 \, + \,2^3c_2 \, + \, ....+ \, (n+1)^3\,C_n$
Now

$(1+x)^n \, = \, C_0 \, + \, C_1x \, + \, C_2x^2 \, + \, ....$
We have the term

$C_2x^2$ whereas we want the term of

$C_2\cdot\, 3^3$ . Hence multiply by x to make it

$3C_2x^3$ . Again differentiate to make

$3^2C_2x^2$ . Again multiply by x to make

$3^2C_2x^2$ and differentiate it to get

$3^2C_2x^2$ and finally put

$x \, = \, 1$ . The above procedure is shown below.
Multiple by x

$x(1+x)^n \, = \, C_0x \, + \, C_1x^2 \, + \, C_3x^3 \, + \, ....$
Differentiate w.r.t.x.

$(1+x)^n \, + \, x.n(1+x)^{n-1} \, = \, C_0 \, + \, 2C_1x \, + \, 3C_2x^2 \, + \, ....$
or

$(1+x)^{n-1} \, [1\, + \,(n+1)x] \, = \,C_0 \, + \, 2C_1x \, + \, 3C_2x^2 \, + \, ....$
Again multiply by x

$(1+x)^{n-1} \, [1\, + \,(n+1)x] \, = \,C_0 \, + \, 2C_1x \, + \, 3C_2x^2 \, + \, ....$
differentiate w.r.t.x.

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

$= \, C_0 \, + \, 2^2C_1x \, + \, 3^2C_2x^2 \, + ....$

$L.H.S. \, = \, (1+x)^{n-2}\,[(n-1)x \, + \, (n^2-1) \, + \, (1+x) \, + \, 2(n+1)\,(x+x^2)]$

$= \, (1+x)^{n-2}\,[1 \, + \, (3n+2)x \, + \, (n^2 + 2n +1)x^2]$

$R.H.S. \, = \, C_0 \, + \, 2^2C_1x \, + \, 3^2C_2x^2 \, + ....$
Again multiply both sides by x and differentiate and lastly put

$x \, = \, 1$

$\sum_{r=0}^{n} \, (r+1)^3 \,C_r \, = \, (n-2)2^{n-3} \, (4+5n+n^2) \, + \, 2^{n-2}\,(8+12n+3n^2)$

If $x^p$ occurs in the expansion of $\left(x^2+\dfrac{1}{x}\right)^{2n}$ , prove that its cofficient is $\dfrac{(2n)!}{\left(\dfrac{1}{3}(4n-p)\right)!\left(\dfrac{1}{3}(2n+p)\right)!}$ .

1973839_1743511_ans_81421dbbdb2649fd8a9920b952303a3c.jpg

Binomial Theorem - Class 11 Commerce Maths - Extra Questions

Find the sum of all rational terms in the expansion of ${ \left( { 3 }^{ 1/5 }+{ 2 }^{ 1/3 } \right) }^{ 15 }$ .

Find the number of terms in the expansion of ${ \left( \sqrt [ 4 ]{ 9 } +\sqrt [ 6 ]{ 8 } \right) }^{ 500 }$ which are integers.

Find the coefficient of $x^5$ in $(x+3)^8$

The coefficient of three consecutive terms in the expansion of ${ \left( 1+a \right) }^{ n }$ are in ratio $1:7:21$ , then find the value of $n$ .

$\dfrac { { C }_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } +\dfrac { { C }_{ 5 } }{ 6 } +...=\dfrac { { 2 }^{ n }-1 }{ n+1 }$

Find the term of the expansion of $(a + b)^{50}$ which is the greatest in absolute value if $\displaystyle\, \left | a \right | = \sqrt{3}\left | b \right |$

Prove that the following equations hold for any natural n.
Prove that $\displaystyle (C_{n}^{o})^2 \, + \, (C_{n}^{1})^2 \, + \, ... \, + \, (C_{n}^{n})^2 \, = \, C_{2n}^{n}.$

Find the power $n$ of the binomial $\displaystyle\, \left ( \frac{x}{5} + \frac{2}{5} \right )^n$ , if the ninth term of the expansion has the greatest coefficient.

Evaluate :
$\left( {\dfrac{{{C_0} + {C_1}}}{{{C_0}}}} \right)\left( {\dfrac{{{C_1} + {C_2}}}{{{C_1}}}} \right)\left( {\dfrac{{{C_2} + {C_3}}}{{{C_2}}}} \right)\left( {\dfrac{{{C_3} + {C_4}}}{{{C_3}}}} \right)........\left( {\dfrac{{{C_{n - 1}} + {C_n}}}{{{C_{n - 1}}}}} \right)$

If in the expansion of $(1+x)^{20}$ , the coefficients of $r^{th}$ and $(r+4)^{th}$ terms are equal, then r is equal to?

If $n> 2$ then prove that ${C}_{1}(a-1)-{C}_{2}\times(a-2)+.....+{(-1)}^{n-1}{C}_{n}(a-n)=a$ ,where ${C}_{r}={ _{ }^{ n }{ C } }_{ r }$

Coefficient of $x^{10}$ in the expansion of $\left(2x^2-\dfrac{3}{x}\right)^{11}, x\neq 0$ .

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

Find a positivevalue of m for which the confficient of $x^{2}$ in the expansion $\left ( 1+x \right )^{m}$ is $6$

The coefficient of ${x}^{20}$ in the expansion of ${(1+{x}^{2})}^{40}({x}^{2}+2+\cfrac{1}{{x}^{2}})$ is

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

In the expansion of $(1+x)^{43}$ the coefficients of the $(2r+1)^{th}$ and the $(r+2)^{th}$ terms are equal; find $r$ .

Find the coefficient of $x^{5}$ in $\left( x + 3 \right)^{n}$

Find the coefficient of $a^{5} b^{7}$ in $\left (a - 2b \right )^{12}$

Find the number of term in the expansion of $(x + a)^{200}+ (x - a)^{200}$ after simplification.

Find the coefficient of $x^{256}$ in $(1 - x)^{101} (x^{2} + x + 1)^{100}$

If $a_n$ denotes the coefficient of $x^n$ in $P(x)=(1+x+2x^2+....+25x^{25})^2$ , find $\dfrac {a_5}{5}$ .

If the sixth term in the expansion of ${ \left( \cfrac { 1 }{ { x }^{ 8/3 } } +{ x }^{ 2 }\log _{ 10 }{ x } \right) }^{ 8 }$ is $5600$ , find $x.$

The value of $p$ , for which coefficient of $x^{50}$ in the expression $(1 + x)^{1000} + 2x (1 + x)^{999} + 3x^{2} (1 + x)^{998} + .... + 1001 x^{1000}$ is equal to $^{1002}C_{p}$ , is

Find $n$ , if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${ \left( \sqrt [ 4 ]{ 2 } +\cfrac { 1 }{ \sqrt [ 4 ]{ 3 } } \right) }^{ n }$ is $\sqrt { 6 } :1$

Let $X = (^{10}C_{1})^{2} + (^{10}C_{2})^{2} + 3(^{10}C_{3})^{2} + .... + 10 (^{10}C_{10})^{2}$ , where $^{10}C_{r}, r\epsilon \left \{1, 2, ..., 10\right \}$ denote binomial coefficients. Then, the value of $\dfrac {1}{1430}X$ is

Evaluate $\sum_{r = 0 }^{n } \, (r \, + \, 1)^3 \, C_r \, whee C_r \, = \, ^nC_r$

If $x^p$ occurs in the expansion of $\left(x^2+\dfrac{1}{x}\right)^{2n}$ , prove that its cofficient is $\dfrac{(2n)!}{\left(\dfrac{1}{3}(4n-p)\right)!\left(\dfrac{1}{3}(2n+p)\right)!}$ .

Class 11 Commerce Maths Extra Questions

Binomial Theorem - Class 11 Commerce Maths - Extra Questions

Find the sum of all rational terms in the expansion of (31/5+21/3)15{ \left( { 3 }^{ 1/5 }+{ 2 }^{ 1/3 } \right) }^{ 15 }.

Find the number of terms in the expansion of (4√9+6√8)500{ \left( \sqrt [ 4 ]{ 9 } +\sqrt [ 6 ]{ 8 } \right) }^{ 500 } which are integers.

Find the coefficient of x5x^5 in (x+3)8(x+3)^8

The coefficient of three consecutive terms in the expansion of (1+a)n{ \left( 1+a \right) }^{ n } are in ratio 1:7:211:7:21, then find the value of nn.

C12+C34+C56+...=2n−1n+1\dfrac { { C }_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } +\dfrac { { C }_{ 5 } }{ 6 } +...=\dfrac { { 2 }^{ n }-1 }{ n+1 }

Find the term of the expansion of (a+b)50(a + b)^{50} which is the greatest in absolute value if |a|=√3|b|\displaystyle\, \left | a \right | = \sqrt{3}\left | b \right |

Prove that the following equations hold for any natural n.Prove that (Con)2+(C1n)2+...+(Cnn)2=Cn2n.\displaystyle (C_{n}^{o})^2 \, + \, (C_{n}^{1})^2 \, + \, ... \, + \, (C_{n}^{n})^2 \, = \, C_{2n}^{n}.

Find the power nn of the binomial (x5+25)n\displaystyle\, \left ( \frac{x}{5} + \frac{2}{5} \right )^n, if the ninth term of the expansion has the greatest coefficient.

If in the expansion of (1+x)20(1+x)^{20}, the coefficients of rthr^{th} and (r+4)th(r+4)^{th} terms are equal, then r is equal to?

If n>2n> 2 then prove that C1(a−1)−C2×(a−2)+.....+(−1)n−1Cn(a−n)=a{C}_{1}(a-1)-{C}_{2}\times(a-2)+.....+{(-1)}^{n-1}{C}_{n}(a-n)=a,where Cr=nCr{C}_{r}={ _{ }^{ n }{ C } }_{ r }

Coefficient of x10x^{10} in the expansion of (2x2−3x)11,x≠0\left(2x^2-\dfrac{3}{x}\right)^{11}, x\neq 0.

Find the middle term in expansion of (2x2−1x)7\left (2x^{2} - \dfrac {1}{x}\right )^{7}.

Find a positivevalue of m for which the confficient of x2x^{2} in the expansion (1+x)m\left ( 1+x \right )^{m} is 66

The coefficient of x20{x}^{20} in the expansion of (1+x2)40(x2+2+1x2){(1+{x}^{2})}^{40}({x}^{2}+2+\cfrac{1}{{x}^{2}}) is

The 2nd2^{nd}, 3rd3^{rd}, 4th4^{th} terms in the expansion of (x+y)n(x+y)^n are 240240, 720720, 10801080 respectively; find x,y,nx, y, n.

In the expansion of (1+x)43(1+x)^{43} the coefficients of the (2r+1)th(2r+1)^{th} and the (r+2)th(r+2)^{th} terms are equal; find rr.

Find the coefficient of x5x^{5} in (x+3)n\left( x + 3 \right)^{n}

Find the coefficient of a5b7a^{5} b^{7} in (a−2b)12\left (a - 2b \right )^{12}

Find the number of term in the expansion of (x+a)200+(x−a)200(x + a)^{200}+ (x - a)^{200} after simplification.

Find the coefficient of x256x^{256} in (1−x)101(x2+x+1)100(1 - x)^{101} (x^{2} + x + 1)^{100}

If ana_n denotes the coefficient of xnx^n in P(x)=(1+x+2x2+....+25x25)2P(x)=(1+x+2x^2+....+25x^{25})^2, find a55\dfrac {a_5}{5}.

If the sixth term in the expansion of (1x8/3+x2log10x)8{ \left( \cfrac { 1 }{ { x }^{ 8/3 } } +{ x }^{ 2 }\log _{ 10 }{ x } \right) }^{ 8 } is 56005600, find x.x.

The value of pp, for which coefficient of x50x^{50} in the expression (1+x)1000+2x(1+x)999+3x2(1+x)998+....+1001x1000(1 + x)^{1000} + 2x (1 + x)^{999} + 3x^{2} (1 + x)^{998} + .... + 1001 x^{1000} is equal to 1002Cp^{1002}C_{p}, is

Find nn, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (4√2+14√3)n{ \left( \sqrt [ 4 ]{ 2 } +\cfrac { 1 }{ \sqrt [ 4 ]{ 3 } } \right) }^{ n } is √6:1\sqrt { 6 } :1

Let X=(10C1)2+(10C2)2+3(10C3)2+....+10(10C10)2X = (^{10}C_{1})^{2} + (^{10}C_{2})^{2} + 3(^{10}C_{3})^{2} + .... + 10 (^{10}C_{10})^{2}, where 10Cr,rϵ{1,2,...,10}^{10}C_{r}, r\epsilon \left \{1, 2, ..., 10\right \} denote binomial coefficients. Then, the value of 11430X\dfrac {1}{1430}X is

Evaluate n∑r=0(r+1)3CrwheeCr=nCr\sum_{r = 0 }^{n } \, (r \, + \, 1)^3 \, C_r \, whee C_r \, = \, ^nC_r

If xpx^p occurs in the expansion of (x2+1x)2n\left(x^2+\dfrac{1}{x}\right)^{2n}, prove that its cofficient is (2n)!(13(4n−p))!(13(2n+p))!\dfrac{(2n)!}{\left(\dfrac{1}{3}(4n-p)\right)!\left(\dfrac{1}{3}(2n+p)\right)!}.

Class 11 Commerce Maths Extra Questions

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Find the sum of all rational terms in the expansion of ${ \left( { 3 }^{ 1/5 }+{ 2 }^{ 1/3 } \right) }^{ 15 }$ .

Find the number of terms in the expansion of ${ \left( \sqrt [ 4 ]{ 9 } +\sqrt [ 6 ]{ 8 } \right) }^{ 500 }$ which are integers.

Find the coefficient of $x^5$ in $(x+3)^8$

The coefficient of three consecutive terms in the expansion of ${ \left( 1+a \right) }^{ n }$ are in ratio $1:7:21$ , then find the value of $n$ .

$\dfrac { { C }_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } +\dfrac { { C }_{ 5 } }{ 6 } +...=\dfrac { { 2 }^{ n }-1 }{ n+1 }$

Find the term of the expansion of $(a + b)^{50}$ which is the greatest in absolute value if $\displaystyle\, \left | a \right | = \sqrt{3}\left | b \right |$

Prove that the following equations hold for any natural n.
Prove that $\displaystyle (C_{n}^{o})^2 \, + \, (C_{n}^{1})^2 \, + \, ... \, + \, (C_{n}^{n})^2 \, = \, C_{2n}^{n}.$

Find the power $n$ of the binomial $\displaystyle\, \left ( \frac{x}{5} + \frac{2}{5} \right )^n$ , if the ninth term of the expansion has the greatest coefficient.

If in the expansion of $(1+x)^{20}$ , the coefficients of $r^{th}$ and $(r+4)^{th}$ terms are equal, then r is equal to?

If $n> 2$ then prove that ${C}_{1}(a-1)-{C}_{2}\times(a-2)+.....+{(-1)}^{n-1}{C}_{n}(a-n)=a$ ,where ${C}_{r}={ _{ }^{ n }{ C } }_{ r }$

Coefficient of $x^{10}$ in the expansion of $\left(2x^2-\dfrac{3}{x}\right)^{11}, x\neq 0$ .

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

Find a positivevalue of m for which the confficient of $x^{2}$ in the expansion $\left ( 1+x \right )^{m}$ is $6$

The coefficient of ${x}^{20}$ in the expansion of ${(1+{x}^{2})}^{40}({x}^{2}+2+\cfrac{1}{{x}^{2}})$ is

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

In the expansion of $(1+x)^{43}$ the coefficients of the $(2r+1)^{th}$ and the $(r+2)^{th}$ terms are equal; find $r$ .

Find the coefficient of $x^{5}$ in $\left( x + 3 \right)^{n}$

Find the coefficient of $a^{5} b^{7}$ in $\left (a - 2b \right )^{12}$

Find the number of term in the expansion of $(x + a)^{200}+ (x - a)^{200}$ after simplification.

Find the coefficient of $x^{256}$ in $(1 - x)^{101} (x^{2} + x + 1)^{100}$

If $a_n$ denotes the coefficient of $x^n$ in $P(x)=(1+x+2x^2+....+25x^{25})^2$ , find $\dfrac {a_5}{5}$ .

If the sixth term in the expansion of ${ \left( \cfrac { 1 }{ { x }^{ 8/3 } } +{ x }^{ 2 }\log _{ 10 }{ x } \right) }^{ 8 }$ is $5600$ , find $x.$

The value of $p$ , for which coefficient of $x^{50}$ in the expression $(1 + x)^{1000} + 2x (1 + x)^{999} + 3x^{2} (1 + x)^{998} + .... + 1001 x^{1000}$ is equal to $^{1002}C_{p}$ , is

Find $n$ , if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of ${ \left( \sqrt [ 4 ]{ 2 } +\cfrac { 1 }{ \sqrt [ 4 ]{ 3 } } \right) }^{ n }$ is $\sqrt { 6 } :1$

Let $X = (^{10}C_{1})^{2} + (^{10}C_{2})^{2} + 3(^{10}C_{3})^{2} + .... + 10 (^{10}C_{10})^{2}$ , where $^{10}C_{r}, r\epsilon \left \{1, 2, ..., 10\right \}$ denote binomial coefficients. Then, the value of $\dfrac {1}{1430}X$ is

Evaluate $\sum_{r = 0 }^{n } \, (r \, + \, 1)^3 \, C_r \, whee C_r \, = \, ^nC_r$

If $x^p$ occurs in the expansion of $\left(x^2+\dfrac{1}{x}\right)^{2n}$ , prove that its cofficient is $\dfrac{(2n)!}{\left(\dfrac{1}{3}(4n-p)\right)!\left(\dfrac{1}{3}(2n+p)\right)!}$ .