MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 12

$C_r \, = \, (^{100 C _ r }) , then \, E = \sum_{r = 0 }^{n + 4 } (-1)^r \, c_r \, c_{r + 1}$

0%
$(^{100} C _ {45} )$
0%
$(^{100} C _ {47} )$
0%
$(^{101} C _ {50} )$
0%
$(^{100 C _ {51} })$

Explanation

$(1 \, + \, x)^{101} \, = \, C_0 \, + \, C_1 x \, + \, C_2 \, x^2 \, + \, .....$

$(1 \, - \, x)^{101} \, = \, C_0 \, x^{101} \, - C_1 \, x^{100} \, + \, C_2 \, x^{99} \, - \, ....$
Multiplying both sides,

$(x^2 \, - \, 1)^{101} \, = \, (C_0 \, + \, C_1 \, x \, + \, C_2 \, x^2 \, + \, ....)$

$(C_0 \, x^{101} \, - \, C_1 \, x^{100} \, + \, C_2 \, x^{99} \, - \, ....)$ ....(1)
Now E =

$C_0 \, C_1 \, - \, C-1 \, C_2 \, + \, C_2 \, C_3 \, - \, ....$
= -[Coefficient of

$x^{100}$ in the product of R.H.S. of (1)]
or = -[Coefficient of

$(x^2)^{50}$ in the product of R.H.S. of (1)] ....(2)
Now

$(x^2 \, - \, 1)^{101} \, = \, (-1)^{101} \, (1 \, - \, x^2)^{101} \, = \, -(1 \, - \, x^2)^{101}$

$\therefore \, \, \,$ Coefficient of

$(x^2)^{50}$
= -

$(- 1)^{50} \, \, ^{101}C_{50} \, = \, - ^{101}C_{50}$ ...(3)

$\therefore \, \, \, \, E \, = \, (-) \, (-) \, ^{101}C_{50} \, = \, ^{101}C_{50} \, by \, (2) \, and \, (3)$

If the

$rth$ term in the expansion of

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

$x^{4}$ then

$r$ is equal to

0%
$2$
0%
$1$
0%
$3$
0%
$5$

$C_{0}$ ,

$C_{1}$ ,

$C_{2}$ ,...,

$C_{n}$ are the binomial coefficients and

$n$ is odd, then

$2C_{1}$ +

$2^{3}.C_{5}+$ ...

$+2^{n}\ ^{n}C_{n}$ equals

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

$\begin{array} { l } { \text { Assertion } ( \mathrm { A } ) : \text { The expansion of } ( 1 + x ) ^ { n } = } \\ { C _ { 0 } + C _ { 1 } x + C _ { 2 } x ^ { 2 } + \ldots + C _ { n } x ^ { n } } \\ { \text { Reason (R): If } x = - 1 , \text { then the above expansion is } } \\ { \text { zero } } \end{array}$

0%
$\begin{array} { l } { \text { Both A and R are true and } \mathrm { R } \text { is the correct } } \\ { \text { explanation of } \mathrm { A } } \end{array}$
0%
$\begin{array} { l } { \text { Both A and R are true and R is not the } } \\ { \text { correct explanation of } \mathrm { A } } \end{array}$
0%
$A$ is true, but $R$ is false
0%
$A$ is false, but $R$ is true

Explanation

We Know that Assertion is nothing but the standard binomial expansion of

$(1+x)^n$

$=C_{0}+C_{1}X+C_{2}X^2+ \dots +C_{n}x^n$

So if

$x=-1$ then

$(1+(-1))^n=(1-1)^n=0$

$Ans:$

$Opt:[B]$

If the ratio

$T_2 : T_3$ in the expansion of

$(a+b)^n$ and

$T_3 : T_4$ in the expansion of

$(a+b)^{n+3}$ are equal , then n

$=$

0%
3
0%
4
0%
5
0%
6

The coefficient of

${x^5}$ in the expansion of

${\left( {1 + {x^2}} \right)^5}{\left( {1 - x} \right)^4}$ is

0%
${4.^6}{C_4}$
0%
${2.^6}{C_4}$
0%
${2.^6}{C_2}$
0%
${4.^6}{C_2}$

In the binomial expansion of

$( a - b ) ^ { n } n > 0$ and the sum

$5 ^ { t h }$ and

$6 ^ { \text { th } }$ terms is zero, then

$\frac { a } { b }$ equal to

0%
$\frac { 5 } { n - 4 }$
0%
$\frac { 6 } { n - 5 }$
0%
$\frac { n - 5 } { 6 }$
0%
$\frac { n - 4 } { 5 }$

The co-efficient of

$x$ in the expansion of

$\left( 1 - 2 x ^ { 3 } + 3 x ^ { 5 } \right) \left( 1 + \dfrac { 1 } { x } \right) ^ { 8 }$ is

0%
$56$
0%
$65$
0%
$154$
0%
$62$

Explanation

$(1-2x^3+3x^5) \left (1+\dfrac {1}{x}\right)^8$ find coefficient of

$x$

$\Rightarrow \left( 1-2{ x }^{ 3 }+3{ x }^{ 5 } \right) \left( 1+8\left( \dfrac { 1 }{ x } \right) +_{ }^{ 8 }{ { C }_{ 2 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 2 }+_{ }^{ 8 }{ { C }_{ 3 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 3 }+_{ }^{ 8 }{ { C }_{ 4 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 4 }+_{ }^{ 8 }{ { C }_{ 5 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 5 }+_{ }^{ 8 }{ { C }_{ 6 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 6 }+_{ }^{ 8 }{ { C }_{ 7 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 7 }+{ \left( \dfrac { 1 }{ x } \right) }^{ 8 } \right)$ (By Binomial theorem)

$=(1+(2.^BC_2 +3^8C_4)x+-----)$

$=$ co-efficient of

$x$ is

$-2\times ^8C_2+3\times ^8C_4=\dfrac {-2\times 8!}{6!\ 2!}+3\times \dfrac {8!}{4!\ 4!}$

$=\dfrac {-2\times 8\times 7}{2\times 1}dfrac {3\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1}=-56+210$

$=154$

If the sum of odd terms and the sum of even terms in

$(x+a)^{n}$ are

$p$ and

$q$ respectively then

$p^{2}+q^{2}=$

0%
$\dfrac{(x+a)^{2n}-(x-a)^{2n}}{2}$
0%
$(x+a)^{2n}-(x-a)^{2n}$
0%
$\dfrac{(x+a)^{2n}+(x-a)^{2n}}{2}$
0%
$(x+a)^{2n}+(x-a)^{2n}$

The sum of the series

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

$n$ is an even positive integer, is equal to

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

Integral part of

$(8+3\sqrt{7})^{n}$ is

0%
an even number
0%
an odd number
0%
an even or odd number depending upon the value of n
0%
nothing can be said

$A$ and

$b$ are coefficients of

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

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

0%
$d = B$
0%
$A = 2 B$
0%
$2 A = B$
0%
$A + B = 0$

The total number of terms in the expansion of

$( x + a ) ^ { 200 } + ( x - a ) ^ { 200 }$ after simplification is

0%
$101$
0%
$102$
0%
$201$
0%
$202$

The sum of the co-efficients of all odd degree terms in the expansion of

${ \left( x+\sqrt { { x }^{ 3 }-1 } \right) }^{ 5 }+{ \left( x-\sqrt { { x }^{ 3 }-1 } \right) }^{ 5 }$

$(x> 1)$ is:

0%
$2$
0%
$-1$
0%
$0$
0%
$1$

Explanation

Sum of the coefficient of odd term is given by

$= 2\left[_{ }^{ 5 }{ { C }_{ 0 }^{ } } + {_{ }^{ 5}}{ { C }_{ 2 }^{ } } + {_{ }^{ 5 }}{ { C }_{ 4 }^{ } }\right]$

$=2\left[1+10-10+5-10+5\right]$

$=2(1+5+5-10)$

$=2$ .

Sum of the coefficient of integral powers of x in

${\left(1-2\sqrt{x}\right)}^{50}$ is

0%
$\dfrac{{3}^{50}+1}{2}$
0%
$\dfrac{{3}^{50}}{2}$
0%
${2}^{49}-1$
0%
${2}^{49}+1$

Find the middle terms(s) in the expansion of

$\left ( 3x-\dfrac{2}{x^{2}} \right )^{15}$ .

0%
$\dfrac {-6435\times 3^7\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^8}{x^9}$
0%
$\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^8}{x^9}$
0%
$\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^7}{x^9}$
0%
$\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^8 \times 2^8}{x^9}$

Sum of coefficients of

${ x }^{ 2r }$ ,

$r= 1,2,3,$ ....... in

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

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

The sum of coefficients of integral powers of

$x$ in the binomial expansion of

$(1-2\sqrt x)^{50}$ is

0%
$\frac{1}{2}(3^{50}-1)$
0%
$\frac{1}{2}(2^{50}+1)$
0%
$\frac{1}{2}(3^{50}+1)$
0%
$\frac{1}{2}(3^{50})$

Coefficient of

$x^{n}$ in expansion of

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

0%
$2n$
0%
$\dfrac{3}{2}(3n^{2}+n)$
0%
$n^{2}+n-1$
0%
$None\ of\ these$

Coefficient of

$x^{r}$ in the expansion of

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

0%
$\dfrac{(2r)!}{(r!)^{2}}$
0%
$\dfrac{(2r)!}{2^{r}(r!)^{2}}$
0%
$\dfrac{(2r)!}{(r!){2^{2r}}}$
0%
$\dfrac{(2r)!}{2^{r}(r+1)!(r+1)}$

Explanation

Coefficient of

${x^r}$ in the expansion of

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

by using binomial expansion

$\begin{array}{l} \Rightarrow \frac { { \left[ { \frac { 1 }{ 2 } \left( { \frac { 1 }{ 2 } +1 } \right) \left( { \frac { 1 }{ 2 } +2 } \right) ...........\left( { \frac { 1 }{ 2 } +r-1 } \right) \left( { 2x } \right) } \right] } }{ { \left( { r! } \right) } } \\ \Rightarrow \frac { { \frac { 1 }{ 2 } .\frac { 3 }{ 2 } .\frac { 5 }{ 2 } .........\left( { r-\frac { 1 }{ 2 } } \right) { 2^{ r } }{ x^{ r } } } }{ { \left( { r! } \right) } } =\frac { { 2r! } }{ { { { \left( { r! } \right) }^{ 2 } } } } \end{array}$

The coefficient of

$x^{99}$ in

$(x+1)(x+3)(x+5).....(x+199)$ is

0%
$1+2+3+...+99$
0%
$1+3+5+...+199$
0%
$1.3.5............199$
0%
$None\ of\ these$

Explanation

$\text x^{99} \text { in }(x+1)(x+3)(x+5) \ldots(x+199) \\$

$\text { Multiplying }(x+1)(x+3) \\$

$\text { will give } x^{2}+(1+3) x+(1)(3) \\$

$\text { so coefficient } x \text { is }(1+3) \\$

$\Rightarrow \operatorname{2n}(x+1)(x+3)(x+5)$

So coefficient of

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

So coefficient of

$x^{n}$ in

$(n+1)$ terms will be sum of coefficients

So we can say it will be sum of roots of equation

So coefficient of

$x^{99}$ will be roots in 100 terms

$1+3+5+\ldots+199$

The sum of coefficient of integral powers of

$x$ in the binomial expansions

${\left( {1 - 2\sqrt x } \right)^{50}}$ is:

0%
$\frac{1}{2}\left( {{3^{50}} - 1} \right)$
0%
$\frac{1}{2}\left( {{2^{50}} + 1} \right)$
0%
$\frac{1}{2}\left( {{3^{50}} + 1} \right)$
0%
$\frac{1}{2}\left( {{3^{50}}} \right)$

The middle term in the expansion of

${\left(3x-\dfrac{{x}^{3}}{6}\right)}^{9}$ is

0%
$\dfrac{21}{16}{x}^{19}$
0%
$\dfrac{-21}{16}{x}^{19}$
0%
$\dfrac{21}{16}{x}^{-19}$
0%
$\dfrac{-21}{16}{x}^{-19}$

The sum of the co-efficient of all odd degree terms in the expansion of

$\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right)$

0%
$0$
0%
$1$
0%
$2$
0%
$- 1$

Sum of the series

$)^{100}C_1)^2 + 2(^{100}C_2)^2 + 3(^{100}C_3)^2+.......+100(^{100}C_{100})^2$ equals

0%
$\frac{2^{99}[1.3.5 ... ... (199)]}{99!}$
0%
$100. ^{200}C_{100}$
0%
$50. ^{200}C_{100}$
0%
$100. ^{199}C_{99}$

The sum of the series

$^{2020}C_0- ^{2020}C_1+^{2020}C_2-^{2020}C_3+.....+^{2020}C_{1010}$ is

0%
$\dfrac{1}{2}^{2020}C_{1010}$
0%
$^{2020}C_{1010}$
0%
Zero
0%
$\dfrac{-1}{2}^{2020}C_{1010}$

Explanation

$(1-x)^{m}={ }^{n} C_{0}-{ }^{n} C_{1} x+\ldots+{ }^{n} C_{n} x^{n}\\$

$\text { Put } x=1 \text { and } n=2020\\$

$0={ }^{2020} C_{0}-{ }^{2020} C_{1}+{ }^{2020} C_{2}+\ldots { }^{2020} C_{2020}\\$

$\text { also }^{n} C_{r}={ }^{n} C_{n-r}\\$

$\text { }^{2020} C_{2020}={ }^{2020} \mathrm{Co}\\$

$\left[2\left[{ }^{2020} \mathrm{C}_{0}-{ }^{2020} \mathrm{C}_{1}+\ldots\right]+{ }^{2020} \mathrm{C}_{1010}\right]=0\\$

$\text { So } { }^{2020} \mathrm{C}_{0}-{ }^{2020} \mathrm{C}_{1}+\ldots-{ }^{2020} \mathrm{C}_{1009}=\\$

$\frac{-1}{2}^{2020} C_{1010}$

The value of

$\dfrac{1}{12!}+\dfrac{1}{10!2!}+\dfrac{1}{8!4!}+...+\dfrac{1}{12!}$

0%
$\dfrac{{2}^{12}}{12!}$
0%
$\dfrac{{2}^{11}}{12!}$
0%
$\dfrac{{2}^{11}}{11!}$
0%
None of these

${ \left( 1+x+2{ x }^{ 2 } \right) }^{ 20 }=0+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+......+{ a }_{ 40 }{ x }^{ 40 }$ then

${ a }_{ 1 }+{ a }_{ 3 }+{ a }_{ 5\quad }+......+{ a }_{ 37\quad }$ equals -

0%
${ 2 }^{ 19}({ 2 }^{ 20 }-21)$
0%
${ 2 }^{ 20 }({ 2 }^{ 19 }-19)$
0%
${ 2 }^{ 19 }({ 2 }^{ 20 }+21)$
0%
None of these

The greatest terms of the expansion

$(2x+5y)^{13}$ when

$x=10$ ,

$y=2$ is?

0%
$^{13}C_5\cdot 20^8\cdot 10^5$
0%
$^{13}C_6\cdot 20^7\cdot 10^4$
0%
$^{13}C_4\cdot 20^9\cdot 10^4$
0%
None of these

The coefficient of x

$^9$ in (x - 1) (x - 4) (x - 9)........(x - 100) is

0%
-235
0%
235
0%
385
0%
None of these

Find the value of

$\dfrac{1}{\left(n-1\right)!}+\dfrac{1}{\left(n-3\right)!3!}+\dfrac{1}{\left(n-5\right)!5!}+...$

0%
$\dfrac{{2}^{n-1}}{(n-1)!}$
0%
$\dfrac{{2}^{n}}{n!}$
0%
$\dfrac{{2}^{n-1}}{n!}$
0%
None of these

The constant term in the expansion of

${ (1+x) }^{ n }{ (1+\frac { 1 }{ x } ) }^{ n }$ is

0%
${ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }+......+(n+1){ C }_{ n }^{ 2 }$
0%
${ ({ C }_{ 0 }+{ C }_{ 1 }+.....+{ C }_{ n }) }^{ 2 }$
0%
${ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+.....+{ C }_{ n }^{ 2 }$
0%
None of these

The largest coefficient in the expansion of

$\left(4+3x\right)^{25}$ is

0%
$^{25}C_{11}3^{25}\left(\dfrac{4}{3}\right)^{14}$
0%
$^{25}C_{11}4^{25}\left(\dfrac{3}{4}\right)^{11}$
0%
$^{25}C_{14}4^{14}3^{11}$
0%
$^{25}C_{14}4^{11}.3^{14}$

Explanation

$=(4+3 x)^{25}\\$

$\left({}^{25} c_{0}(4)+{ }^{25} c_{1}(4)^{2}(3 x)+\cdots^{25} c_{25}\right.\\$

$(4+3 x)^{25}\\$

$\text { Let Tr the term is largest }\\$

$\therefore \quad T_{r-1}<T_{r}>T_{r+1}\\$

$=\frac{T_{2}}{T_{r_{-1}}}>1 \quad \frac{T_{r+1}}{T_{r}}<1$

$T_{r+1^{}}= {}^{25} C_{r}(4)^{25-r}(3 x)^{k}$

$T_{r-1}=25_{C_{r-2}}(4)^{25-r+2}(3 x)^{r-2}$

$T_{r}={ }^{25} C_{r-1}\left(4\right)^{26-r}(3 x)^{r-1}$

$\therefore \frac{25 C_{r-1}(4)^{26-r}(3 x)^{r-1}}{25 C_{r-2}(4)^{27-r}(3 x)^{r-2}}>1$

$\frac{(25-r)}{(x-1)} \times \frac{3}{4}>1$

${}75-3 r > 4 r -4\\$

$79>7 r\\$

$11.1>r$

$\Rightarrow r=12$

largest term =

$\left({ }^{25}c_{11}(3)^{25}\left(\frac{4}{3}\right)^{14}\right.)$

The sum of the co-efficient of all odd degree terms in the expansion of

$\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } , ( x > 1 )$ is :

0%
$- 1$
0%
$1$
0%
$0$
0%
$2$

If the last term in the binomial expansion of

$\left(2^{1/3}-\dfrac {1}{\sqrt {2}}\right)^{n}$ is

$\left(\dfrac {1}{3^{5/3}}\right)^{\log_{3}8}$ , then the

$5^{th}$ terms form the beginning is:

0%
$210$
0%
$420$
0%
$103$
0%
$None\ of\ these$

Coefficient of

$x^{25}$ in

$(1+x+x^{2}+x^{3}+....+x^{10})^{7}$ is

0%
$^{31}C_{15}-7.^{20}C_{14}$
0%
$^{31}C_{14}-7.^{20}C_{14}$
0%
$31$
0%
$None\ of\ these$

The coefficient of

$x^{8}$ in

$(1+2x^{2}-x^{3})^{9}$ is

0%
$1680$
0%
$2140$
0%
$2520$
0%
$2730$

The coefficients of

$x^{10}$ in the expansion of

$(1+x)^{15}+(1+x)^{16}+(1+x)^{17}+....+(1+x)^{30}$ is

0%
$^{31}C_{10}-^{15}C_{10}$
0%
$^{31}C_{11}-^{15}C_{11}$
0%
$^{30}C_{10}-^{15}C_{10}$
0%
$^{31}C_{10}-^{14}C_{11}$

The sum of the coefficient in the expansion of

$(a+2b+c)^{11}$ is-

0%
$4^{11}$
0%
$32$
0%
$31$
0%
$None\ of\ these$

State true or false.

The general term for

$3,7,13,21,31,43$ ........ is

$n^2−(n−1),n=1,2,3,...$

0%
True
0%
False

The coefficient of

$x^{3}$ in the expansion of

$(1+2x+3x^{2})^{10}$ is

0%
Less than $200$
0%
Less than $400$ but greater than $200$
0%
$1400$
0%
$1500$

The coefficient of

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

$\dfrac { 1 }{ \left( 1-x \right) \left( 1-2x \right) \left( 1-3x \right) }$ is

0%
$\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 3 }^{ n+3 }+1 \right)$
0%
$\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 2 }^{ n+3 }+1 \right)$
0%
$\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 3 }^{ n+2 }+1 \right)$
0%
$None\ of\ these$

Explanation

$\text { We know that } \\$

$\frac{1}{1-a} \text { is sum of } \infty GP=1+a+a^{2}+a^{3} \ldots\infty \\$

$\text { Similary: }$

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

${\left(1+(3 x)+(3 x)^{2}+(3 x)^{3}\ldots\infty\right.)}$

From this equation, we observe that

coefficient of

$x^{n}$ is

$: \rightarrow$

$\frac{1}{2}\left[2^{n+2}-3^{n+3}+1\right]$

Option (a)

The number of rational terms in the expansion of

${ \left( 1+\sqrt { 2 } +\sqrt [ 3 ]{ 3 } \right) }^{ 6 }$ is

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

The coefficient of

$x^4$ in the expansion of

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

0%
$990$
0%
$495$
0%
$330$
0%
none of these

Explanation

$=\left(1+x+x^{2}+x^{3}\right)^{11}$

$=\left((1+x)\left(1+x^{2}\right)\right)^{11}$

$=(1+x)^{ 11}\left[1+^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots^{11} c_{11}\left(x^{2}\right)^{11}\right.]$

$=\left[1+{ }^{11} c_{1} x+{ }^{11} c_{2} x^{2}+\cdots\right]\left[1+{ }^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots\right]$

$={ }^{11} c_{0} \cdot{ }^{11} c_{2} \cdot x^{4}+{ }^{11} c_{2} \cdot{ }^{11} c_{1} \cdot x^{4}+{ }^{11} c_{4}{ }^{11} c_{0} x^{4}$

$=\quad 990 x^{4}$

Option

$(A)$

The coefficient of x

$^{24}$ in the expansion of
(1 +3x + 6x

$^2$ + 10x

$^3$ + -----------+

$\infty$ )

$^{2/3}$ =

0%
300
0%
250
0%
25
0%
205

Explanation

$\left(1+3 x+6 x^{2}+10 x^{3}+\ldots\infty\right)^{\frac{2}{3}} \\$

$\text { We know that, } \\$

${ (1+\alpha) }^{n}=1+n \alpha+\frac{n(n-1) \alpha^{2}}{2 !}+\ldots \\$

$\text { comparing with given eqn, }$

$\frac{n(n-1)}{2} \cdot \alpha^{2}=6 x^{2} \\$

$=n(n-1) \frac{9 x^{2}}{n^{2}}=6 x^{2} \\$

$\Rightarrow n=-3$

$n \alpha=3 x$

$\Rightarrow\alpha=\frac{3 x}{n}$

$\alpha=-x$

$\therefore \text {Expansion is} (1-x)^{-\not 3 \times \frac{2}{\not3}}$

$\quad =(1-x)^{-2}$

$=1+2 x+\frac{2 \times 3}{2 !} x^{2}+\frac{2 \times 3 \times 4}{3 !} x^{3}+\ldots$

$\therefore$ coeff. of

$x^{24}$ is 2

$= \frac{2 \times 3 \times 4 \times\ldots25}{24 !}$

$=25$

The co-efficient of

$x^{k}$ in expansion of

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

$\left(n>k\right)$

0%
$^{n}C_{k}$
0%
$^{n+1}C_{k}$
0%
$^{n+1}C_{k+1}$
0%
$None\ of\ these$

The number of terms in the expansion of

${ \left[ { a }^{ 3 }+\dfrac { 1 }{ { a }^{ 3 } } +1 \right] }^{ 100 }$ is

0%
$201$
0%
$300$
0%
$200$
0%
$^{ 100 }{ C }_{ 3 }$

Explanation

$\left(a^{3}+\frac{1}{a^{3}}+1\right)^{100} \\$

$\frac{1}{a^{300}}\left(1+a^{3}+a^{6}\right)^{100}$

Here, terms will be in the form,

$a^{0} ; a^{3}, a^{6}\cdots$

and highest power term will be

$\left(a^{6}\right)^{100}$

This forms a A.P of powers,

$\therefore \quad 600=0+(n-1) 3$

$\therefore \quad n=201 \\$

$\text { Option (a) }$

For

$x\in R, x\neq -1$ if

$(1+x)^{2016}+x(1+x)^{2015}+x(1+x)^{2014}+.+x^{2016}=\sum _{ i=0 }^{ 2016 }{ { a }_{ i }{ x }^{ i } }$ , then

$a_{17}$ is equal to

0%
$\dfrac{2017!}{ 17!2000!}$
0%
$\dfrac{2016!}{ 17!1999!}$
0%
$\dfrac{2017!}{2000!}$
0%
$\dfrac{2016!}{ 16!}$

The middle term in the expansion of

$\left(1-\dfrac{1}{x}\right)^{n}\left(1-x\right)^{n}$ is

0%
$^{2n}C_{n}$
0%
$-^{2n}C_{n}$
0%
$-^{2n}C_{n-1}$
0%
$none\ of\ these$

If the middle term in the expansion of

$( 1 + x ) ^ { 2 n }$ is the greatest term, then

$x$ lies in the interval ___________________.

0%
$\left( \frac { n } { n + 1 } , \frac { n + 1 } { n } \right)$
0%
$\left( \frac { n + 1 } { n } , \frac { n } { n + 1 } \right)$
0%
$( n - 2 , n )$
0%
$( n - 1 , n )$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 12 - 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 12 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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