If $$f(n)=\sum _{ k=1 }^{ n }{ \sum _{ j=k }^{ n }{ ({ _{ }^{ n }{ C } }_{ j } } ) } ({ _{ }^{ j }{ C } }_{ k })$$, find $$f(n)$$.

$${ 3 }^{ n-1 }-{ 2 }^{ n -1}$$

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

MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 5

In the expansion of

$\displaystyle \left ( \sqrt{2} + \sqrt[5]{3}\right )^{120}$ the number of irrational terms is

0%
$12$
0%
$13$
0%
$108$
0%
$54$

Explanation

Total number of rational terms is

$\dfrac{120}{L.C.M(5,2)}+1$

$=\dfrac{120}{10}+1$

$=13$
Hence total number of irrational terms are

$=121-13$

$=108$

If the fourth term in the expansion of

$\displaystyle { \left( \sqrt { \frac { 1 }{ { x }^{ \log { x } +1 } } } +{ x }^{ 1/12 } \right) }^{ 6 }$ is equal to

$200$ and

$x>1$ , then

$x$ is equal to

0%
${ 10 }^{ \sqrt { 2 } }$
0%
$10$
0%
${10}^{4}$
0%
None of these

Explanation

Given

${ T }_{ 4 }=200$

$\displaystyle \Rightarrow ^{ 6 }{ { C }_{ 3 } }{ \left( \sqrt { \frac { 1 }{ { x }^{ \log _{ 10 }{ x+1 } } } } \right) }^{ 3 }{ \left( { x }^{ \frac { 1 }{ 12 } } \right) }^{ 3 }=200$

$\displaystyle \Rightarrow 20\times { x }^{ \frac { 3 }{ 2\left( \log _{ 10 }{ x+1 } \right) } +\frac { 1 }{ 4 } }=200\Rightarrow { x }^{ \left( \frac { 3 }{ 2\left( \log _{ 10 }{ x+1 } \right) } +\frac { 1 }{ 4 } \right) }=10$

$\displaystyle \Rightarrow \left( \frac { 3 }{ 2\left( \log _{ 10 }{ x+1 } \right) } +\frac { 1 }{ 4 } \right) =\log _{ x }{ 10 } =\frac { 1 }{ \log _{ 10 }{ x } }$

$\displaystyle \Rightarrow \frac { 3 }{ 2\left( y+1 \right) } +\frac { 1 }{ 4 } =\frac { 1 }{ y }$ where

$y=\log _{ 10 }{ x }$

$\Rightarrow y=-4$ or

$y=1$

$\Rightarrow \log _{ 10 }{ x } =-4$ or

$1$

$\Rightarrow x={ 10 }^{ -4 }$ or

$10$

$\Rightarrow x=10\quad \left( \because x>1 \right)$

$A$ and

$B$ are coefficients of

${x}^{n}$ in the expansions of

${ (1+x) }^{ 2n }$ and

${ (1+x) }^{ 2n-1 }$ respectively, then

$\dfrac AB$ is equal to

0%
$1$
0%
$2$
0%
$\dfrac 12$
0%
$\dfrac 1n$

Explanation

We know that coefficient of

${x}^{r}$ in the expansion of

${ (1+x) }^{ m }$ is

${ _{ }^{ m }{ C } }_{ r }$
Thus

$A={ _{ }^{ 2n }{ C } }_{ n }\quad ;\quad B={ _{ }^{ 2n-1 }{ C } }_{ n }$
We have

$\cfrac { A }{ B } =\cfrac { { _{ }^{ 2n }{ C } }_{ n } }{ { _{ }^{ 2n-1 }{ C } }_{ n } } =\cfrac { (2n)! }{ n!n! } \cfrac { (n!)(n-1)! }{ (2n-1)! } =\cfrac { 2n }{ n } =2$

The expression

${ C }_{ 0 }+2{ C }_{ 1 }+3{ C }_{ 2 }+......+(n+1){ C }_{ n }$ is equal to

0%
${ 2 }^{ n-1 }$
0%
$n({ 2 }^{ n-1 })$
0%
$n({ 2 }^{ n-1 })+{ 2 }^{ n }$
0%
$(n+1){ 2 }^{ n }$

Explanation

Let

$E={ C }_{ 0 }+2{ C }_{ 1 }+3{ C }_{ 2 }+......n{ C }_{ n-1 }+(n+1){ C }_{ n }\quad ........(1)$
Using

${ C }_{ r }={ C }_{ n-r }\quad E=(n+1){ C }_{ 0 }+n{ C }_{ 1 }+(n-1){ C }_{ 2 }+.....+2{ C }_{ n-1 }+{ C }_{ n }\quad ........(2)$
Adding

$(1)$ and

$(2)$ , we get

$2E=(n+2){ C }_{ 0 }+(n+2){ C }_{ 1 }+(n+2){ C }_{ 2 }+....+(n+2){ C }_{ n }$

$=(n+2)\left\{ { C }_{ 0 }+{ C }_{ 1 }+.....+{ C }_{ n } \right\} =(n+2){ 2 }^{ n }$

$\Rightarrow \quad E=(n+2){ 2 }^{ n-1 }$

$f(n)=\sum _{ k=1 }^{ n }{ \sum _{ j=k }^{ n }{ ({ _{ }^{ n }{ C } }_{ j } } ) } ({ _{ }^{ j }{ C } }_{ k })$ , find

$f(n)$ .

0%
${ 3 }^{ n }-{ 2 }^{ n }$
0%
${ 3 }^{ n }+{ 2 }^{ n }$
0%
${ 3 }^{ n-1 }-{ 2 }^{ n -1}$
0%
${ 3 }^{ n+1 }-{ 2 }^{ n+1 }$

Explanation

Simplifying the above expression we get

$\:^{n}C_{1}(\:^{1}C_{1})+\:^{n}C_{2}(\:^{2}C_{1}+\:^{2}C_{2})+...+\:^{n}C_{n}(\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n})$

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

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

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

$=3^{n}-2^{n}$ .

If the third term in the expansion

${ \left( x+{ x }^{ \log _{ 5 }{ x } } \right) }^{ 5 }$ is

$2$ , then

$x$ equals

0%
$1/5,5$
0%
$1/5,1/\sqrt {5}$
0%
$\sqrt{5},5$
0%
$1/\sqrt{5},5$

Explanation

$T_{2+1}$

$=\:^{5}C_{2}x^{3+2log_{5}(x)}$

$=10x^{3+2log_{5}(x)}$

$=2$

$x^{3+2log_{5}(x)}=\dfrac{1}{5}$

$log_{5}(x)(3+2log_{5}(x))=-1$

$2log(x)^{2}+3log(x)+1=0$

$2log(x)^{2}+2log(x)+log(x)+1=0$

$(2log(x)+1)(log(x)+1)=0$

$log(x)=-1$

$x=\dfrac{1}{5}$ and

$log(x)=\dfrac{-1}{2}$

$x=\dfrac{1}{\sqrt{5}}$

${ C }_{ r }=\begin{pmatrix} n \\ r \end{pmatrix}$ , then sum of series

${ \dfrac{{C }_{ 0 }^{ 2 }}{1}}+\dfrac{{ C }_{ 1 }^{ 2 }}{2}+\dfrac{{ C }_{ 2 }^{ 2 }}{3}+....$ upto

$(n+1)$ terms is

0%
$\cfrac { 1 }{ n+1 } \begin{pmatrix} 2n+1 \\ n+1 \end{pmatrix}$
0%
$\cfrac { 1 }{ 2(n+1) } \begin{pmatrix} 2n+1 \\ n+1 \end{pmatrix}$
0%
$\cfrac { 1 }{ n+1 } \begin{pmatrix} 2n+1 \\ n \end{pmatrix}$
0%
$\cfrac { (2n+1)! }{ { (n+1)! }^{ 2 } }$

Explanation

For

$r\ge 0$

$\cfrac { 1 }{ r+1 } { C }_{ r }^{ 2 }=\cfrac { 1 }{ r+1 } \cdot \cfrac { n! }{ r!(n-r)! } { C }_{ r }$

$=\cfrac { 1 }{ n+1 } \cfrac { (n+1)! }{ (r+1)!(n-r)! } { C }_{ r }\quad$

$=\cfrac { 1 }{ n+1 } \begin{pmatrix} n+1 \\ n-r \end{pmatrix}\begin{pmatrix} n \\ r \end{pmatrix}$
Thus,

$S=\cfrac { 1 }{ n+1 } { S }_{ 1 }$ where

${ S }_{ 1 }=\sum _{ r=0 }^{ n }{ \begin{pmatrix} n+1 \\ n-r \end{pmatrix}\begin{pmatrix} n \\ r \end{pmatrix}\quad }$

$=$ number of ways of selecting

$n$ persons out of

$(n+1)$ men and

$n$ women

$=$ number of ways of selecting

$n$ persons out of

$(2n+1)$ persons

$=\begin{pmatrix} 2n+1 \\ n+1 \end{pmatrix}=\cfrac { (2n+1)! }{ n!(n+1)! }$

$\therefore$

$S=\cfrac { 1 }{ n+1 } \cfrac { (2n+1)! }{ n!(n+1)! } =\cfrac { 1 }{ (n+1) } \begin{pmatrix} 2n+1 \\ n+1 \end{pmatrix} = \cfrac{1}{n+1}\begin{pmatrix}2n+1\\n \end{pmatrix} = \cfrac{1}{n+1}\cfrac{(2n+1)!}{n!(n+1)!} = \cfrac{(2n+1)!}{(n+1)!^2}$

Hence, options

$A,C$ and

$D$ are correct.

Value of

$S=1\times 2\times 3\times 4+2\times 3\times 4\times 5+.......+n(n-1)(n+2)(n+3)$ is

0%
$\cfrac { 1 }{ 5 } n(n+1)(n+2)(n+3)(n+4)$
0%
$\cfrac { 1 }{ 5! } ({ _{ }^{ n+3 }{ C } }_{ 5 })$
0%
$\cfrac { 1 }{ 5 }{ _{ }^{ n+4 }{ C } }_{ 4 }$
0%
None of these

Given that the

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

${ \left( 2+\cfrac { 3x }{ 8 } \right) }^{ 10 }$ has the maximum numerical value, then

$x$ can lie in the interval(s)

0%
$\left( 2,\cfrac { 64 }{ 21 } \right)$
0%
$\left( -\cfrac { 60 }{ 23 } ,-2 \right)$
0%
$\left( -\cfrac { 64 }{ 21 } ,-2 \right)$
0%
$\left( 2,-\cfrac { 60 }{ 23 } \right)$

Explanation

Since

${t}_{4}$ is numerically the greatest term,

$\left| { t }_{ 3 } \right| <\left| { t }_{ 4 } \right| ;\quad \left| { t }_{ 5 } \right| <\left| { t }_{ 4 } \right|$

$\quad \Rightarrow \left| \cfrac { { t }_{ 3 } }{ { t }_{ 4 } } \right| <1;\quad \left| \cfrac { { t }_{ 5 } }{ { t }_{ 4 } } \right| <1$
but

$\cfrac { { t }_{ 3 } }{ { t }_{ 4 } } =\cfrac { { _{ }^{ 10 }{ C } }_{ 2 }({ 2 }^{ 8 }){ \left( \dfrac {3x}{8} \right) }^{ 2 } }{ { _{ }^{ 10 }{ C } }_{ 3 }({ 2 }^{ 7 }){ \left( \dfrac {3x}{8} \right) }^{ 3 } } =\cfrac { 2 }{ x }$
and

$\cfrac { { t }_{ 5 } }{ { t }_{ 4 } } =\cfrac { { _{ }^{ 10 }{ C } }_{ 4 }({ 2 }^{ 6 }){ \left( \dfrac {3x}{8} \right) }^{ 4 } }{ { _{ }^{ 10 }{ C } }_{ 3 }({ 2 }^{ 7 }){ \left( \dfrac {3x}{8} \right) }^{ 3 } } =\cfrac { 21x }{ 64 }$
Now,

$\left| \cfrac { { t }_{ 3 } }{ { t }_{ 4 } } \right| <1\Rightarrow \quad \left| x \right| >2$
and

$\left| \cfrac { { t }_{ 5 } }{ { t }_{ 4 } } \right| <1\Rightarrow \quad \left| x \right| <\cfrac { 64 }{ 21 }$

If the third term in the expansion of

${ \left( \cfrac { 1 }{ x } +{ x }^{ \log _{ 2 }{ x } } \right) }^{ 5 }$ is

$40$ then

$x$ equals

0%
$\dfrac {1}{\sqrt{2}},2$
0%
$\sqrt{2},4$
0%
$\dfrac {1}{\sqrt{2}},4$
0%
$\sqrt{2},\dfrac{1}{\sqrt{2}}$

Explanation

The middle term of

$\displaystyle { \left( \frac { 1 }{ x } +x\log _{ 2 }{ x } \right) }^{ 5 }$ is

$\displaystyle ^{ 5 }{ { C }_{ 2 } }{ \left( \frac { 1 }{ x } \right) }^{ 3 }{ \left( { x }^{ \log _{ 2 }{ x } } \right) }^{ 2 }=10{ x }^{ 2\log _{ 2 }{ x-3 } }=40$

$\therefore{ x }^{2 \log _{ 2 }{ x-3 } }=4={ 2 }^{ 2 }$

Taking logarithm to the base

$2$ of both the sides, we get

$(2t-3)t=2$ where

$t=\log _{ 2 }{ x }$

$\Rightarrow 2{ t }^{ 2 }-3t-2=0$ or

$\left( 2t+1 \right) \left( t-2 \right) =0$

$\displaystyle \Rightarrow t=-\frac { 1 }{ 2 } , 2\Rightarrow x=\frac { 1 }{ \sqrt { 2 } },4$

If in the expansion of

${ \left( \cfrac { 1 }{ x } +x\tan { x } \right) }^{ 5 }$ the ratio to 4th term to the 2nd term is

$\cfrac { 2 }{ 27 } { \pi }^{ 4 }$ , then the value of

$x$ can be

0%
$\cfrac { -\pi }{ 6 }$
0%
$\cfrac { -\pi }{ 3 }$
0%
$\cfrac { \pi }{ 3 }$
0%
$\cfrac { \pi }{ 12 }$

Explanation

$T_{4}=\dfrac{2\pi^{4}}{27}T_{2}$

$\:^{5}C_{3}x \tan^{3}x=\dfrac{2\pi^{4}}{27}\:^{5}C_{1}\dfrac{1}{x^{3}}tanx$

$10x^{4} \tan^{2}(x)=5\dfrac{2\pi^{4}}{27}$

$x^{4} \tan^{2}(x)=\dfrac{\pi^{4}}{27}$

$x^{4} \tan^{2}(x)=\dfrac{\pi^{4}}{81}.\sqrt{3}^{2}$
Hence

$x^{4}=\dfrac{\pi^{4}}{81}$

$x=\pm\dfrac{\pi}{3}$

$\tan^{2}(x)=\sqrt{3}^{2}$

$\tan(x)=\pm\sqrt{3}$

$x=\pm\dfrac{\pi}{3}$
Hence verified.

If the coefficients of 2nd, 3rd and the 4th terms in the expansion of

${ \left( 1+x \right) }^{ n }$ are in A.P, then value of

$n$ is

0%
$5$
0%
$7$
0%
$11$
0%
$14$

Explanation

For the terms to be in A.P, it must follow

$(n-2r)^{2}=n+2$
In the above case

$r=2$
Substituting in the equation, we get

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

$n^{2}-8n+16=n+2$

$n^{2}-9n+14=0$

$(n-2)(n-7)=0$

$n=2$ and

$n=7$
However for

$n=2$ we will have only

$3$ terms.
Hence the required answer is

$7.$

If the middle term of

${ \left( x+\cfrac { 1 }{ x } \sin ^{ -1 }{ x } \right) }^{ 8 }$ is equal to

$\dfrac {630}{16}$ , then value of

$x$ is (are)

0%
$\dfrac {\pi}3$
0%
$\dfrac {\pi}6$
0%
$-\dfrac {\pi}6$
0%
$-\dfrac {\pi}3$

Explanation

Middle term

$={ _{ }^{ 8 }{ C } }_{ 4 }{ x }^{ 4 }{ \left( \cfrac { 1 }{ x } \sin ^{ -1 }{ x } \right) }^{ 4 }=70{ \left( \sin ^{ -1 }{ x } \right) }^{ 4 }$

$70{ \left( \sin ^{ -1 }{ x } \right) }^{ 4 }=\dfrac{630}{16}$

$\Rightarrow {\left(\sin ^{-1}{x} \right)}^4=\dfrac 9{16}$

$\Rightarrow \sin ^{ -1 }{ x } =\pm \dfrac {\sqrt { 3 }}{ 2} \Rightarrow x=\pm \dfrac{\pi }3$

Let

${ S }_{ n }= \sum _{ r=0 }^{ n }{ { (-2) }^{ r } } \left( \cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ r+2 }{ C } }_{ r } } \right)$ , then

0%
${ S }_{ n }=\cfrac { 1 }{ n+1 }$ if $n$ is odd
0%
${ S }_{ n }=\cfrac { 1 }{ n+2 }$ if $n$ is odd
0%
${ S }_{ n }=\cfrac { 1 }{ n+1 }$ if $n$ is even
0%
${ S }_{ n }=\cfrac { 1 }{ n+2 }$ if $n$ is even

Explanation

$\cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ r+2 }{ C } }_{ r } } =\cfrac { !n }{ r!(n-r)! } \cfrac { r!2! }{ (r+2)! } =\cfrac { 2(n!) }{ (r+2)!(n-r)! }$

$\cfrac { 2 }{ (n+1)(n+2) } \cfrac { (n+2)! }{ (r+2)!\left[ (n+2)-(r+2) \right] ! } =\cfrac { 2 }{ (n+1)(n+2) } ={ _{ }^{ n+2 }{ C } }_{ r+2 }$
Thus,

$S=\sum _{ r=0 }^{ n }{ { (-2) }^{ r } } \left( \cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ r+2 }{ C } }_{ r } } \right) \quad =\cfrac { 2 }{ (n+1)(n+2) } \sum _{ r=0 }^{ n }{ { (-2) }^{ r } } { _{ }^{ n+2 }{ C } }_{ r+2 }\quad$
Putting

$r+2=s$

$=\cfrac { 2 }{ 4(n+1)(n+2) } \sum _{ s=2 }^{ n+2 }{ { _{ }^{ n+2 }{ C } }_{ s } } { (-2) }^{ s }\quad$

$=\cfrac { 1 }{ 2(n+1)(n+2) } \times \left[ \sum _{ s=0 }^{ n+2 }{ { _{ }^{ n+2 }{ C } }_{ s } } { (-2) }^{ s }-{ _{ }^{ n+2 }{ C } }_{ 0 }{ (-2) }^{ 0 }-{ _{ }^{ n+2 }{ C } }_{ 1 }{ (-2) }^{ 1 } \right]$

$=\cfrac { 1 }{ 2(n+1)(n+2) } \left[ { (1-2) }^{ n+2 }-1+2(n+2) \right] =\cfrac { 1 }{ 2(n+1)(n+2) } \left[ 2n+3+{ (-1) }^{ n } \right]$
But

$\quad 2n+3+{ (-1) }^{ n }=\{ 2(n+2)$ if

$n$ is even

$2(n+1)$ if

$n$ is odd

$\}$
Thus

$S=\{ \cfrac { 1 }{ n+1 }$ , if

$n$ is even

$\cfrac { 1 }{ n+2 }$ , if

$n$ is odd

$\}$

Let

${C}_{r}$ stand for

${ _{ }^{ n }{ C } }_{ r }$ and

$S(n,r)={C}_{0}-{C}_{1}+{C}_{2}-{C}_{3}+....+{ (-1) }^{ r }{ C }_{ r }$

0%
If $S(n,r)=28$ then $n=9,r=2$ or $n=9,r=6$
0%
If $S(n,r)=-15$ , then $n=7,r=3$
0%
$S(n,n)=0$
0%
none of these

Explanation

$S(n,r)=$ coefficient of

${x}^{r}$ in

$\left( 1+x+{ x }^{ 2 }+... \right) \left( { C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-...+{ \left( -1 \right) }^{ n }{ C }_{ n }{ x }^{ n } \right)$

$=$ Coefficient of

${x}^{r}$ in

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

Now,

${ \left( -1 \right) }^{ r }\left( ^{ n-1 }{ { C }_{ r } } \right) =28$

This is possible if

$n=9,r=2$ or

$n=9, r=6$

$S(n,n)=\sum_{n=0}^{n} (-1)^n\ { }^{n}C_{n}=0$

Hence, option A and C are correct.

Value of

$\sum _{ k=1 }^{ \infty } \sum _{ r=0 }^{ k }{ \cfrac { 1 }{ { 3 }^{ k } } } ({ _{ }^{ k }{ C } }_{ r })$ is

0%
$2$
0%
$\cfrac{2}{3}$
0%
$\cfrac{1}{3}$
0%
none of these

Explanation

$\sum \dfrac{1}{3^{k}}\:^{k}C_{r}$

$=\dfrac{1}{3^{k}}\sum\:^{k}C_{r}$

$=\dfrac{2^{k}}{3^{k}}$
This is a G.P
Therefore, the sum of the series will be

$S=\dfrac{\frac{2}{3}}{1-\frac{2}{3}}$

$=2$

Value of

$\sum { \sum _{ 0\le r<s\le n }^{ }{ { (r+s)\left( { C }_{ r }+{ C }_{ s } \right) }^{ 2 } } }$ is

0%
$n\left[ (n-1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right]$
0%
$n\left[ (n+1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right]$
0%
$n\left[ { 2 }^{ 2n }-n({ _{ }^{ 2n }{ C } }_{ n }) \right]$
0%
None of these

Explanation

$S=\sum _{ 0\le r }^{ }{ \sum _{ <s\le n }^{ }{ { (n-r+n-s)({ C }_{ n-r }-{ C }_{ n-s }) }^{ 2 } } }$

$=2n\sum _{ 0\le r }^{ }{ \sum _{ <s\le n }^{ }{ { ({ C }_{ r }+{ C }_{ s }) }^{ 2 } } } -S$

$\Rightarrow S=n\left[ (n-1)({ _{ }^{ 2n }{ C } }_{ n })+{ 2 }^{ 2n } \right]$

$A=^{ 2n }{ { C }_{ 0 } }.^{ 2n }{ { C }_{ 1 } }+^{ 2n }{ { C }_{ 1 } }^{ 2n-1 }{ { C }_{ 1 } }+^{ 2n }{ { C }_{ 2 } }^{ 2n-2 }{ { C }_{ 1 } }+...$ then

$A$ is

0%
$0$
0%
$n.2^{2n}$
0%
${ 2 }^{ 10 }-2$
0%
$1$

Explanation

Given

$A=^{2n}C_0.^{2n}C_1+^{2n}C_1.^{2n-1}C_1+^{2n}C_2.^{2n-2}C_1+........$

we can write

$A$ as summation .

$\Rightarrow A=\sum_{r=0}^{2n} .^{2n}C_r.^{2n-r}C_1=\sum_{r=0}^{2n}.\dfrac{(2n)!}{r!(2n-r)!}.\dfrac{(2n-r)!}{(2n-r-1)!1!}=\sum_{r=0}^{2n}\dfrac{(2n)!}{r!(2n-r-1)!}=2n\sum_{r=0}^{2n}.^{2n-1}C_r$

$= 2n.2^{2n-1}=n.2^{2n}$ .

$\therefore A=n.2^{2n}$

The coefficient of

${ x }^{ 9 }$ in the expansion of

${ \left( { x }^{ 3 }+\cfrac { 1 }{ { 2 }^{ t } } \right) }^{ 11 }$ , where

$t=\log _{ \sqrt { 2 } }{ { (x }^{ \tfrac 32 } } )$ ,

0%
$-5$
0%
$330$
0%
$520$
0%
$5+\log _{ \sqrt { 2 } }{ (3 } )$

Explanation

$2^{ t }= { 2 }^{\displaystyle \log _{ \sqrt { 2 } }{ { x }^{\tfrac 32 } } }= { 2 }^{\displaystyle \log _{ 2 }{ { x }^{ 3 } } } = x^{ 3 }$
The general term in the expansion of

$\left( x^{ 3 }+\displaystyle\frac { 1 }{ x^{ 3 } } \right) ^{ 11 }$ is

$T_{ r+1 }= ^{ 11 }C_{ r }\left( x^{ 3 } \right) ^{ 11-r }\displaystyle\frac { 1 }{ x^{ 3r } } = ^{ 11 }C_{ r }x^{ 33-6r }$
To find the coefficient of

$x^{ 9 }$
Hence,

$33-6r = 9$

$\Rightarrow r= 4$
Hence, the coefficient of

$x^{ 9 }$ is

$^{ 11 }C_{ 4 } = 330$ .

Value of

$\sum { \sum _{ 0\le i<j\le n }^{ }{ { \left( { C }_{ i }+{ C }_{ j } \right) }^{ 2 } } }$ is

0%
${ 2 }^{ 2n }(n+1)({ _{ }^{ 2n }{ C } }_{ n })$
0%
${ 2 }^{ 2n }+(n+1)({ _{ }^{ 2n }{ C } }_{ n })$
0%
${ 2 }^{ 2n-1 }+(n+1)({ _{ }^{ 2n }{ C } }_{ n })$
0%
none of these

Explanation

$\displaystyle\sum _{ 0\le i }^{ }{ \sum _{ \le j\le n }^{ }{ { \left( { C }_{ i }+{ C }_{ j } \right) }^{ 2 } } } i=0,1,2,...,\left( n-1 \right)$ where

$j=1,2,3,...,n$ and

$i<j$

$\displaystyle=n\left( { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+...+{ C }_{ n }^{ 2 } \right) +2\sum { \sum { { C }_{ i }{ C }_{ j }0\le i<j\le n } }$

$\displaystyle=n.^{ 2n }{ { C }_{ n } }+\left[ { \left( { C }_{ 0 }+{ C }_{ 1 }+...+{ C }_{ n } \right) }^{ 2 }-\left( { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+...+{ C }_{ n }^{ 2 } \right) \right]$

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

${x}^{2r}$ occurs in

${ \left( x+\cfrac { 2 }{ { x }^{ 2 } } \right) }^{ n }$ , then

$n-2r$ must be of the form

0%
$3k$
0%
$3k-1$
0%
$3k+1$
0%
$4k\pm 1$

Explanation

Writing the general term

$T_{k+1}=\:^{n}C_{r}x^{n-3k}2^{k}$
For

$x^{2r}$

$n-3k=2r$

$n-2r=3k$
Where K is a positive integer

$\geq 0$

Value(s) of

$x$ for which the fourth term in the expansion of

${ \left( { \sqrt { x } }^{ 1/(\log _{ 2 }{ x+1 } ) }+{ x }^{ 1/2 } \right) }^{ 6 }$ is

$40$ is (are)

0%
$1/8$
0%
$2$
0%
$1/16,2$
0%
$1/8,4$

Explanation

${ t }_{ 4 }={ _{ }^{ 6 }{ C } }_{ 3 }\left[ { x }^{ 1/2(\log _{ 2 }{ x } +1) } \right] { \left( \cfrac { 1 }{ { x }^{ 12 } } \right) }^{ 3 }=40\quad$

$\Rightarrow \left( \cfrac { 3 }{ 2t+2 } +\cfrac { 1 }{ 4 } \right) t=1$ where

$t=\log _{ 2 }{ x }$

$\quad \Rightarrow \quad 7t+{ t }^{ 2 }=4t+4\quad \Rightarrow \quad t=-4,1$

$\quad \therefore \quad x=1/16,2$

If the middle term of

${ \left( x+\cfrac { 1 }{ x } \sin ^{ -1 }{ x } \right) }^{ 8 }$ is

$\cfrac { 35{ \pi }^{ 4 } }{ 8 }$ , then value of

$x$ can be

0%
$\dfrac12$
0%
$\dfrac{\sqrt {3}}2$
0%
$\dfrac1{\sqrt {2}}$
0%
$1$

Explanation

Middle term is

$T_{5}$

$T_5=\ ^{8}C_{4}(\sin^{-1}(x))^{4}$

$=70(\sin^{-1}(x))^{4}$

$=\dfrac{35}{8}\pi^{4}$

Hence,

$70(\sin^{-1}(x))^{4}=\dfrac{70}{16}\pi^{4}$

$\Rightarrow(\sin^{-1}(x))^{4}=\dfrac{\pi^{4}}{16}$

$\Rightarrow(\sin^{-1}(x)=\pm\dfrac{\pi}{2}$

$\Rightarrow x=\pm1$

The number of irrational terms in the expansion of

${ \left( { 4 }^{ 1/5 }+{ 7 }^{ 1/10 } \right) }^{ 45 }$ is

0%
40
0%
5
0%
41
0%
none of these

Explanation

Total number of terms in the expansion of

$\displaystyle { \left( { 4 }^{ \frac { 1 }{ 5 } }+{ 7 }^{ \frac { 1 }{ 10 } } \right) }^{ 45 }$ is

$45+1=46$

The general term in the expansion is

$\displaystyle { T }_{ r+1 }=^{ 45 }{ { C }_{ r } }\times { 4 }^{ \frac { 45-r }{ 5 } }\times { 7 }^{ \frac { r }{ 10 } }$

${ T }_{ r+1 }$ is rational if

$r=0,10,20,30,40$

$\therefore$ Number of rational terms

$=5$

$\therefore$ Number of irrational terms

$=46-5=41$

If the expansion of

${ \left( 1+x \right) }^{ 50 }$ , the sum of coefficients of add powers of

$x$ is

0%
${ 2 }^{ 50 }$
0%
${ 2 }^{ 49 }$
0%
$0$
0%
None of these

Explanation

The sum of coefficients of odd powers of

$x$

$=_{ }^{ 50 }{ { C }_{ 1 }^{ } }+^{ 50 }{ { C }_{ 3 }^{ } }+...+^{ 50 }{ { C }_{ 49 }^{ } }={ 2 }^{ 50-1 }={ 2 }^{ 49 }$ .

The coefficients of three consecutive terms in the expansion of

${ (1+x) }^{ n }$ are in the ratio

$5:10:21$ . find

$n$

0%
$5$
0%
$4$
0%
$7$
0%
$8$

The coefficient of

$\displaystyle x^{99}$ in the polynomial

$\displaystyle \left ( x-1 \right )\left ( x-2 \right )\left ( x-3 \right )...\left ( x-100 \right )$ is

0%
$1$
0%
$-5050$
0%
$5050$
0%
$-1$

Explanation

If we multiply the brackets and take

$x$ from each bracket except one, that is, we take

$x$ from

$99$ brackets and the constant term from the last bracket we will have

$x^99$

So coefficient of

$x^99$ will be sum of all the constant terms taken once in this manner.

Coefficient of

$x^{99}=-\left( 1+2+3+....+100 \right) =\dfrac { -100\left( 101 \right) }{ 2 } =-5050$

Value of the expression

$\quad { C }_{ 0 }+({ C }_{ 0 }+{ C }_{ 1 })+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 })+....+({ C }_{ 0 }+{ C }_{ 1 }+....+{ C }_{ n-1 })$ is

0%
${2}^{n-1}$
0%
$n{2}^{n-2}$
0%
$n{2}^{n-1}$
0%
$2n$

Explanation

Simplifying, we get

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

$=n[\:^{n}C_{0}+\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}]-[\:^{n}C_{1}+2\:^{n}C_{2}+3\:^{n}C_{3}+...n\:^{n}C_{n}]$

$=n2^{n}-\cfrac{d(1+x)^{n}}{dx}|_{x=1}$

$=n2^{n}-n2^{n-1}$

$=n2^{n-1}$

${ S }_{ n }=\cfrac { 1 }{ n+1 } \sum _{ i=0 }^{ n }{ \begin{pmatrix} n \\ i \end{pmatrix} }$ , then

$2{ S }_{ n+1 }-{ S }_{ n }$ equals

0%
$=2^{n}[\dfrac{3n-2}{(n+1)(n+2)}]$
0%
$=2^{n}[\dfrac{3n+2}{(n+1)(n+2)}]$
0%
$=2^{n}[\dfrac{3n+2}{(n-1)(n+2)}]$
0%
0

Explanation

$\sum \:^{n}C_{i}=2^{n}$ ....sum of binomial coefficients of

$(1+x)^{n}$

Hence

$S_{n}=\dfrac{2^{n}}{n+1}$

Now

$S_{n+1}=\dfrac{2^{n+1}}{n+2}$

$2S_{n+1}-S_{n}$

$=\dfrac{2^{n+2}}{n+2}-\dfrac{2^{n}}{n+1}$

$=2^{n}[\dfrac{4}{n+2}-\dfrac{1}{n+1}]$

$=2^{n}[\dfrac{4n+4-n-2}{(n+1)(n+2)}]$

$=2^{n}[\dfrac{3n+2}{(n+1)(n+2)}]$

In the expansion of

$\displaystyle \left ( x^{3}+3.2^{-\log_{2}x^{3}} \right )^{11}$

0%
there appears a term with the power $\displaystyle x^{2}$
0%
there does not appear a term with the power $\displaystyle x^{2}$
0%
there appear a term with the power $\displaystyle x^{-3}$
0%
the ratio of the co-efficient of $\displaystyle x^{3}$ to that of $\displaystyle x^{-3}$ is $\displaystyle \frac{1}{3}$

Explanation

$\displaystyle \left ( x^{3}+3.2^{-\log_{2}x^{3}} \right )^{11}=\left ( x^{3}+\frac{3}{x^{3}} \right )^{11}$

$\displaystyle T_{r+1}=^{11}C_{r}(x^{3})^{11-r}\left ( \frac{3}{x^{3}} \right )^{r}=\:^{11}C_{r}(x)^{33-6r}(3)^{r}$
Now

$33 - 6r = 2$

$\displaystyle \Rightarrow$

$6r = 31$ (not possible)

$33 - 6r = -3$

$\displaystyle \Rightarrow$

$r = 6$

$33 - 6r = 3$

$\displaystyle \Rightarrow$

$r = 5$

$\displaystyle \therefore \frac{coeff.of\:\:x^{3}}{coeff. of\:\:x^{-3}}=\frac{^{11}C_{5}3^{5}}{^{11}C_{6}3^{6}}=\frac{1}{3}$

$\displaystyle \left ( 1+x+x^{2} \right )^{25}=a_{0}+a_{1}x+a_{2}x^{2}+\cdot \cdot \cdot +a_{50}\cdot x^{50}$ then

$\displaystyle a_{0}+a_{2}+a_{4}+\cdot \cdot \cdot +a_{50}$ is

0%
even
0%
odd & of the form $3n$ .
0%
odd & of the form $(3n-1)$
0%
Odd & of the form $(3n+1)$

Explanation

$(1+x+x^{2})^{25}=a_{0}+a_{1}x+a_{2}x^{2}+...a_{50}x^{50}$

Substituting

$x=1$ , we get

$3^{25}=a_{0}+a_{1}+a_{2}+...a_{50}$ ...(i)

Substituting

$x=-1$ we get

$1=a_{0}-a_{1}+a_{2}-a_{3}...+a_{50}$ ...(ii)

Adding

$i$ and

$ii$ , we get

$3^{25}+1=2[a_{0}+a_{2}+a_{4}+...a_{50}]$

Hence

$a_{0}+a_{2}+a_{4}+...a_{50}=\dfrac{3^{25}+1}{2}$

Since the LHS is a positive integer hence the RHS has to be a positive integer.
Therefore

$3^{25}+1$ will be divisible by

$2$ .

Hence

$3^{25}+1$ will be even.

And

$3^{2n+1}+1$ is always even and is divisible by

$4$ .

Hence

$\dfrac{3^{25}+1}{2}$ will be even.

$\displaystyle \begin{pmatrix} p \\ q \end{pmatrix}$ =0 for

$\displaystyle p< q$ p,

$\displaystyle q\epsilon W$ then

$\displaystyle \sum_{r=0}^{\infty}$

$\displaystyle \begin{pmatrix} n \\ 2r \end{pmatrix}$

0%
$\displaystyle 2^{n}$
0%
$\displaystyle 2^{n-1}$
0%
$\displaystyle 2^{2n-1}$
0%
$\displaystyle \hat{2n}C_{n}$

Explanation

$\sum \:^{n}C_{2r}$
Is the sum of even odd term in the binomial expansion of

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

$\sum \:^{n}C_{2r}$ will always be equal to

$2^{n-1}$ .

Number of terms free from radical sign in the expansion of

$\displaystyle \left ( 1+3^{1/3}+7^{1/2} \right )^{10}$ is

0%
4
0%
5
0%
6
0%
8

Explanation

$(1+(3^{\frac{1}{3}}+7^{\frac{1}{2}}))^{10}$
Let

$(3^{\frac{1}{3}}+7^{\frac{1}{2}})=x$
Then the above expression reduces to,

$(1+x)^{10}$
Now writing the general term, we get

$T_{r+1}=\:^{10}C_{r}x^{r}$

$=\:^{10}C_{r}(3^{\frac{1}{3}}+7^{\frac{1}{2}})^{r}$ .
Hence we get one rational term for each

$r=0,2,3,4,6,8,9,10$
Hence in total 8 terms.

The sum of the coefficients of all the even powers of x in the expansion of

$\displaystyle \left ( 2x^{2}-3x+1 \right )^{11}$ is

0%
$\displaystyle 2 . 6^{10}$
0%
$\displaystyle 3 . 6^{10}$
0%
$\displaystyle 6^{11}$
0%
None of the above

Explanation

Given equation is

$(2x^{2}-3x+1)^{11}$

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

$=(3)^{11}.(2)^{11-1}$

$=3^{11}.2^{10}$

$=6^{10}.3$

Number of rational terms in the expansion of

$\displaystyle \left ( \sqrt{2}+\sqrt[4]{3} \right )^{100}$ is

0%
$25$
0%
$26$
0%
$27$
0%
$28$

Explanation

$\textbf{term of the binomial expansion}$

Since, ${r^{th}}$ term of the expression ${\left( {a + b} \right)^n}$ is ${T_{r + 1}} = {}^n{C_r}{\left( a \right)^{n - r}}{\left( b \right)^r}$

Therefore, ${r^{th}}$ term of the expression ${\left( {\sqrt 2 + \sqrt[4]{3}} \right)^{100}}$ is ${T_{r + 1}} = {}^{100}{C_r}{\left( 2 \right)^{\frac{{100 - r}}{2}}}{\left( 3 \right)^{\frac{r}{4}}}$

$\textbf{Step 2 : Apply the condition of a number to be a rational number}$

${r^{th}}$ term of the expression is a rational term if

$\frac{{100 - r}}{2}$ and $\frac{r}{4}$ should be integer

Therefore, $r$ should be multiple of 4 $\Rightarrow r = 0,4,8,12.............100$

Which forms an A.P. with first term $(a) = 0;$ common difference $(d) = 4;$

${n^{th}}$ term $({A_n}) = 100$

since, $A_{n}=a+(n-1)d$

$\Rightarrow 100 = 0 + (n - 1)4$

$\Rightarrow n = 26$

$\textbf{Hence, Option ‘B’ is correct.}$

The sum of the binomial coefficients of

$\displaystyle \left [2 x+\frac{1}{x} \right ]^{n}$ is equal to value of n is

0%
5
0%
6
0%
7
0%
8

Explanation

Substituting

$x=1$ for sum of coefficients, we get

$3^{n}=243$

$3^{n}=3^{5}$
Hence

$n=5$

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

$(x + a)^n$ will be

0%
$^nC_rx^na^{n-r}$
0%
$^nC_rx^{n-r}a^{r}$
0%
$^nC_rx^{n-r}a^{n}$
0%
$^nC_rx^ra^{n-r}$

Explanation

$\\ { (x+a) }^{ n }={ x }^{ n }+{ C }_{ 1 }^{ n }{ x }^{ n-1 }a+...+{ C }_{ r }^{ n }{ x }^{ n-r }{ a }^{ r }+...{ a }^{ n }\\ \quad the\quad (r+1)th\quad term\quad is\quad { C }_{ r }^{ n }{ x }^{ n-r }{ a }^{ r }.\quad Hence\quad option\quad (B)\\$

If coefficients of

$x^n$ in

$(1+x)^{101}(1-x+x^2)^{100}$ is non-zero then

$n$ cannot be of the form

0%
$3t+1$
0%
$3t$
0%
$3t+2$
0%
$4t+1$

Explanation

Given expression is

$(1+x)^{101} (1-x+x^{2})^{100}$ , Let it be

$y$

$y = (1+x)((1+x)(1-x+x^{2}))^{100}=(1+x)(1+x^{3})^{100}$

We know that

$(1+x^{3})^{100}=1+ax^{3}+bx^{6}+cx^{9}+....$

$\Rightarrow y= (1+x)(1+x^{3})^{100}=1+x+ax^{3}+ax^{4}+bx^{6}+bx^{7}+....$

We can observe that

$x^{3t+2}$ terms are missing , which means the coefficients of

$x^{3t+2}$ terms are

$0$

Therefore for the coefficient of

$x^{n}$ not to be zero ,

$n$ cannot be

$3t+2$

$\displaystyle \binom{n}{o}+\binom{n}{1}+\binom{n}{2}+.........+\binom{n}{n}=$

0%
$\displaystyle 2^{n-1}$
0%
$\displaystyle ^{2n}C_{n}$
0%
$\displaystyle 2^n$
0%
$\displaystyle 2^{n+1}$

Explanation

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

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

Find the

$\displaystyle 7^{th}$ term of

$\displaystyle \left ( 3x^{2}-\frac{1}{3}\right)^{10}$ .

0%
$\displaystyle \frac{66}{3}x^{7}$
0%
$\displaystyle \frac{70}{3}x^{7}$
0%
$\displaystyle \frac{66}{3}x^{8}$
0%
$\displaystyle \frac{70}{3}x^{8}$

Explanation

Writing the general term, we get

$T_{r+1}=(-1)^{r}\:^{10}C_{r}3^{10-2r}.x^{20-2r}$

For 7th term r is equal to 6.

Hence

$T_{7}=(-1)^{6}\:^{10}C_{6}3^{10-12}.x^{20-12}$

$=\dfrac{210.x^{8}}{9}$

$=\dfrac{70x^{8}}{3}$ .

The value of

$\displaystyle 2\sum_{r=0}^{n}a_{2r-1}$ is

0%
$\displaystyle 9^{n}-1$
0%
$\displaystyle 9^{n}+1$
0%
$\displaystyle 9^{n}-2$
0%
$\displaystyle 9^{n}+2$

Explanation

$(1+4x+4x^{2})^{n}=a_{0}+a_{1}x+a_{2}x^{2}+...(a_{2n}x^{2n})$
Substituting

$x=1$ we get

$9^{n}=a_{0}+a_{1}+a_{2}+...(a_{2n})$
Substituting

$x=-1$ we get

$1=a_{0}-a_{1}+a_{2}-a_{3}...(a_{2n})$
Adding both we get

$2(a_{0}+a_{2}+a_{4}+...a_{2n})=9^{n}+1$
Hence

$2\sum _{k=0} ^{n}a_{2k}=9^{n}+1$

${C}_{r}$ stands for

$^{n}{C}_{r}$ , then the sum of the series

$\displaystyle\frac { 2\left( \frac { n }{ 2 } \right) !\left( \frac { n }{ 2 } \right) ! }{ n! } \left[ { C }_{ 0 }^{ 2 }-2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }-...+{ \left( -1 \right) }^{ n }\left( n+1 \right) { C }_{ n }^{ 2 } \right]$ where

$n$ is an even positive integer, is

0%
$0$
0%
${ \left( -1 \right) }^{ n/2 }\left( n+1 \right)$
0%
${ \left( -1 \right) }^{ n/2 }\left( n+2 \right)$
0%
${ \left( -1 \right) }^{ n }n$

Explanation

$\displaystyle{ C }_{ 0 }^{ 2 }-2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }-4{ C }_{ 3 }^{ 2 }+...+{ \left( -1 \right) }^{ n }\left( n+1 \right) { C }_{ n }^{ 2 }$

$\displaystyle=\left[ { C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }-{ C }_{ 3 }^{ 2 }+...+{ \left( -1 \right) }^{ n }{ C }_{ n }^{ 2 } \right] -\left[ { C }_{ 1 }^{ 2 }-2{ C }_{ 2 }^{ 2 }+3{ C }_{ 3 }^{ 2 }-...+{ \left( -1 \right) }^{ n }n{ C }_{ n }^{ 2 } \right]$

$\displaystyle={ \left( -1 \right) }^{ \frac { n }{ 2 } }\frac { n! }{ \left( \frac { n }{ 2 } \right) !\left( \frac { n }{ 2 } \right) ! } -{ \left( -1 \right) }^{ \frac { n }{ 2 } -1 }\frac { n }{ 2 } .^{ n }{ { C }_{ \frac { n }{ 2 } } }$

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

$\displaystyle\therefore 2\left( \frac { n }{ 2 } \right) !\left( \frac { n }{ 2 } \right) !\left[ { C }_{ 0 }^{ 2 }-2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }-...+{ \left( -1 \right) }^{ n }\left( n+2 \right) { C }_{ n }^{ 2 } \right]$

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

Find the middle terms in the expansion of

$\displaystyle \left ( 2x^{2}-\frac{1}{x} \right )^{7}$

0%
$\displaystyle -560x^{5},\:280x^{2}$
0%
$\displaystyle -280x^{5},\:560x^{2}$
0%
$\displaystyle 560x^{5},\:-280x^{2}$
0%
$\displaystyle 280x^{5},\:-560x^{2}$

Explanation

The middle terms will be

$T_{4}$ and

$T_{5}$
Now,

$T_{4}$

$=T_{3+1}$

$=-\:^{7}C_{3}.2^{4}.x^{8-3}$

$=-35.(16).x^{5}$

$=-560.x^{5}$

$T_{5}$

$=T_{4+1}$

$=\:^{7}C_{4}.2^{3}.x^{6-4}$

$=35.(8).x^{2}$

$=280.x^{2}$

The number of terms in the expansion of

$( 1 + 5\sqrt 2 x)^9 + ( 1 -5\sqrt 2 x)^9$ is

0%
5
0%
7
0%
9
0%
10

Explanation

$S=(1+5\sqrt{2}x)^{9}+(1-5\sqrt{2}x)^{9}$
Let

$k=5\sqrt{2}$
Then the above binomial expansion changes to

$S=(1+k)^{9}+(1-k)^{9}$

$=[1+\:^{9}C_{1}k+\:^{9}C_{2}k^{2}+...\:^{9}C_{9}k^{9}]+[1-\:^{9}C_{1}k+\:^{9}C_{2}k^{2}-\:^{9}C_{3}k^{3}...-\:^{9}C_{9}k^{9}]$

$=2[1+\:^{9}C_{2}k^{2}+\:^{9}C_{4}k^{4}+\:^{9}C_{6}k^{6}+\:^{9}C_{8}k^{8}]$
Hence in total 5 terms will be there.

In the expansion of

$(1 + x)^n$ , the sum of coefficients of odd powers of

$x$ is

0%
$2^n+1$
0%
$2^n-1$
0%
$2^n$
0%
$2^{n-1}$

Explanation

$\left( 1+x \right) ^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }+...+{ C }_{ n }{ x }^{ n }$

Putting

$x=1$ and

$x=1$ and subtracting, we get.

${ 2 }^{ n }=2\left( { C }_{ 1 }+{ C }_{ 3 }+{ C }_{ 5 }+... \right)$

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

Or the sum of the coefficients of the odd power of

$x$ is

${2}^{n-1}$

$(1+x)^n=C_0+C_1x+C_2x^2+.....+C_nx^2$ , then the value of

$C_0+C_2+C_4+ .....$ is

0%
$2^{n-1}$
0%
$2^n-1$
0%
$2^n$
0%
$2^{n-1}-1$

Explanation

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

$x=1$

$2^{n}=\:^{n}C_{0}+\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}$ ...(i)
Let

$x=-1$

$0=\:^{n}C_{0}-\:^{n}C_{1}+\:^{n}C_{2}+...(-1)^{n}\:^{n}C_{n}$ ...(ii)
Adding both gives us

$2^{n}=2[\:^{n}C_{0}+\:^{n}C_{2}+\:^{n}C_{4}+...]$

$2^{n=1}=\:^{n}C_{0}+\:^{n}C_{2}+\:^{n}C_{4}+...$
Hence the answer is

$2^{n-1}$

Which term in the expansion of

$\left (\displaystyle \frac {x}{3}-\frac {2}{x^2}\right )^{10}$ contains

$x^4$ ?

0%
1
0%
3
0%
4
0%
5

Explanation

The general term is as follows

$T_{r+1}$

$=(-1)^{r}\:^{10}C_{r}x^{10-3r}3^{10-r}2^{r}$
Hence, for the coefficient of

$x^{4}$

$10-3r=4$

$3r=6$

$r=2$ .

So,

$r+1=3$ rd term

$C_0, C_1, C_2, ..............C_n$ are binomial coefficients then

$\displaystyle \frac {1}{n!0!}+\frac {1}{(n-1)!1!}+\frac {1}{(n-2)!2!}+ ....+\frac {1}{0!n!}$ is equal to

0%
$2^n$
0%
$\displaystyle \frac {2^{n-1}}{n!}$
0%
$\displaystyle \frac {2^n}{n!}$
0%
none of these

Explanation

Multiplying and dividing the entire expression by n!, we get

$\dfrac{1}{n!}[\dfrac{n!}{0!.n!}+\dfrac{n!}{(n-1)!.1!}+\dfrac{n!}{(n-2)!.2!}+...\dfrac{n!}{0!.n!}]$

$=\dfrac{1}{n!}[\:^{n}C_{0}+\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n}]$

$=\dfrac{1}{n!}(1+x)^{n}_{x=1}$

$=\dfrac{2^{n}}{n!}$ .

If the sum of the coefficients in the expansion of

$(1 -3x + 10x^2)^n$ is

$a$ and if the sum of the coefficients in the expansion of

$(1 + x^2)^n$ is

$b$ , then

0%
$a = 3b$
0%
$a = b^3$
0%
$b=a^3$
0%
none of these

Explanation

Given ,

$a =$ sum of the coefficient in the expansion of

$(1 - 3x + 10x^2)^n$

$\Rightarrow a = (1 - 3 + 10)^n = (8)^n$ (Putting

$x = 1$ )

$\Rightarrow a = (2)^{3n}$
Now,

$b =$ sum of the coefficients in the expansion of

$(1 + x^2)^n$

$\Rightarrow b = (1 + 1)^n = 2^n$ (Putting

$x = 1$ )
Clearly,

$a = b^3$

$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+ .... +2^n\binom{n}{n}$ is equal to

0%
$2^n$
0%
$0$
0%
$3^n$
0%
none of these

Explanation

Given,

$\binom{n}{0}+2\binom{n}{1}+2^2\binom{n}{2}+ .... +2^n\binom{n}{n}$

Simplifying the above expression we get

$=\:^{n}C_{0}2^{0}+\:^{n}C_{1}2^{1}+\:^{n}C_{2}2^{2}+...\:^{n}C_{n}2^{n}$
Consider

$x=2$ in the expansion of

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

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

$=3^{n}$ .

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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