MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 9

In the expansion of

$(1+x)^{30}$ , the sum of the coefficients of odd powers of

$x$ , is

0%
$2^{30}$
0%
$2^{31}$
0%
$0$
0%
$2^{29}$

The coefficient of

$x^{10}$ in the expansion of

$(1+x)^{2}(1+x^{2})^{3}(1+x^{3})^{4}$ is qual to:

0%
$52$
0%
$56$
0%
$50$
0%
$44$

Explanation

We know,

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

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

$(a + b)^4 = a^4 +4a^3b + 6a^2b^2 + 4ab^3 + b^4$

$(1+x)^2(1+x^2)^3(1+x^3)^4$

$\Rightarrow$

$(1+x^2+2x)(1+3x^2+3x^4+x^6)(1+4x^3+6x^6+4x^9+x^{12})$

$\Rightarrow$

$(1+3x^2+3x^4+x^6+x^2+3x^4+3x^6+x^8+2x+6x^3+6x^5+2x^7)(1+4x^3+6x^6+4x^9+x^{12})$

$\Rightarrow$

$(1+2x+4x^2+6x^4+6x^5+4x^6+2x^7+x^8)(1+4x^3+6x^6+4x^9+x^{12})$

Now, coefficient of

$x^{10}$

$=(4\times 2)+(6\times 6)+(2\times 4)$

$=8+36+8$

$=52$

$P$ be the sum of odd terms and

$Q$ be the sum of even terms in the expansion of

$(x+a)^n$ , then

$P ^ { 2 } - Q ^ { 2 } = \left( x ^ { 2 } - a ^ { 2 } \right) ^ { n }$

0%
True
0%
False

$(1+x-2x^2)^8=1+a_1x+a_2x^2+........ + a_{16}x^{16}$ , then

$a_1+a_3+a_5+......a_{15}$ =

0%
$2^7$
0%
$-2^7$
0%
$3^2$
0%
$4^6$

The sum of coefficients of the two middle terms in the expansion of

$( 1 + x ) ^ { 2 n - 1 }$ is equal to

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

Explanation

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

$2n$ terms.

Hence there are

$2$ middle terms, the

$n^{th}$ term and the

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

$t_n=(2n-1)C_{n-1}\cdot x^{n-1}$

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

Sum of coefficients

$=(2n-1)C_{n-1}+(2n-1)C_n=2nC_n$ .

1214006_1259665_ans_fea019b1d6ca4e3ebd83a2db59644afe.jpg

$^{2007}C_0-8.^{2007}C_1+13.^{2007}C_2-18.^{2007}C_3+....$ upto 2008 terms =

0%
$0$
0%
$2007$
0%
$-2007$
0%
$2008$

If the coefficients of

$x^{-7}$ and

$x^{-8}$ to the expansion of

$\left(2+\dfrac {1}{3x}\right)^{n}$ are equal then

$n=$

0%
$56$
0%
$15$
0%
$45$
0%
$55$

The coefficient of

$x^{4}$ in the expansion of

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

0%
$\dfrac{405}{256}$
0%
$\dfrac{504}{259}$
0%
$\dfrac{450}{263}$
0%
$none\ of\ these$

The coefficient of

$x^{5}$ in the expansion of

$(1+x)^{21}+(1+x)^{22}+....+(1+x)^{30}$ is

0%
$^{51}C_{5}$
0%
$^{9}C_{5}$
0%
$^{31}C_{6}-^{21}C_{6}$
0%
$^{30}C_{5}-^{20}C_{5}$

Find the middle term in the expansion of

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

0%
$252$
0%
$248$
0%
$230$
0%
$200$

The coefficient of

${x}^{3}$ in the expression of

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

0%
$10$
0%
$220$
0%
$211$
0%
$None \ of \ these$

$(1+ax)^n=1+8x+24x^2+.....$ then

0%
$a=3$
0%
$a=4$
0%
$a=2$
0%
$a=5$

The coefficient of

$x^{n}$ in the expansion of

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

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

Explanation

We have

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

$= \dfrac{1}{{2\left( {1 - x} \right)}} - \dfrac{4}{{1 - 2x}} + \dfrac{9}{{2\left( {1 - 3x} \right)}}$

$= \dfrac{1}{2}{\left( {1 - x} \right)^{ - 1}} - 4{\left( {1 - 2x} \right)^{ - 1}} + \dfrac{9}{2}{\left( {1 - 3x} \right)^{ - 1}}$

$= \dfrac{1}{2}\left( {1 + x + {x^2} + ..... + {x^n}} \right) - 4\left( {1 + 2x + {{\left( {2x} \right)}^2} + ..... + {{\left( {2x} \right)}^n}} \right) + \dfrac{9}{2}\left( {1 + 3x + {{\left( {3x} \right)}^2} + .... + {{\left( {3x} \right)}^n}} \right)$

Coefficient of

${x^n}$ , we get

$= \dfrac{1}{2}\left[ {1 - {{8.2}^n} + {{9.3}^n}} \right]$

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

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

Hence, the option

$(B)$ is correct answer.

If the

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

$\left(\dfrac {1}{x^{8/3}}+x^{2}\log ^{x}_{10}\right)^{8}$ is

$5600$ , then equals

0%
$1$
0%
$\log^{10}_{e}$
0%
$10$
0%
$No\ such\ x\ exists$

The number of terms in

$\left(x^{3}+1+\dfrac{1}{x^{3}}\right)^{100}$ is

0%
$300$
0%
$200$
0%
$100$
0%
$201$

The coefficient of

$x^{4}$ in the expansion

$\left(1+5x+9x^{2}+.+\left(4k+1\right)x^{k}+..\right)\left(1+x^{2}\right)^{11}$ is

0%
$^{11}C_{2}+4 ^{11}C_{1}+3$
0%
$^{11}C_{2}+3 ^{11}C_{1}+4$
0%
$3 ^{11}C_{2}+4 ^{11}C_{1}+3$
0%
$171$

If the coefficients of

$2^{nd}, 3^{rd},$ and

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

${(1 + x)^n},n \in N$ are in A.P.,then

$n=$

0%
$7$
0%
$14$
0%
$2$
0%
None of these

The coefficient of

$x^8$ in the expansion of

$(1+x+x^3+x^5+x^9)(1+x^2)^5(1_x^4)^6$ is equal to

0%
180
0%
100
0%
80
0%
None of these

In the expansion of

$\displaystyle {\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{1/3}} + 1}} - \frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}$ , the term which does not contain

$x$ is -

0%
$^{11}{C_4}{ - ^{10}}{C_3}$
0%
$^{10}{C_7}$
0%
$^{10}{C_4}$
0%
$^{11}{C_5}{ - ^{10}}{C_5}$

The coefficient of

${ x }^{ n }$ in

$(1-x+\frac { { x }^{ 2 } }{ 2! } -\frac { { x }^{ 3 } }{ 3! } +.....+\frac { { (-1) }^{ n }{ x }^{ n } }{ n! } )^{ 2 }$ is

0%
$\frac { { (-n) }^{ n } }{ n! }$
0%
$\frac { { (-2) }^{ n } }{ n! }$
0%
$\frac { 1 }{ (n!)^{ 2 } }$
0%
$-\frac { 1 }{ (n!)^{ 2 } }$

If coefficient of

$2^{nd}, 3^{rd}$ and

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

$\left(1+x\right)^{2n}$ are in A.P. then :

0%
$2n^{2}+9n+7=0$
0%
$2n^{2}-9n+7=0$
0%
$2n^{2}-9n-7=0$
0%
$2n^{2}+9n-7=0$

Explanation

$^{2n}C_{1},^{2n}C_{2}, ^{2n}C_{3}$ are in A.P

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

$=2\times (\frac{2n(2n-1)}{1\times 2})=\frac{2n}{1}+\frac{2n(2n-1)(2n-2)}{1\times 2\times 3}$

$=2n-1=1+\frac{(2n-1)(n-1)}{3}$

$= 6n-3=3n+2n^{2}-3n+1$

$\Rightarrow 2n^{2}-9n+7=0$

1198100_1296832_ans_24bafc178b1845bbb58b77213426566b.jpg

The coefficient of

$a^3b^4c$ in the expansion of

$(1+a+b-c)^9$ is

0%
$2.^9C_7.^7C_4$
0%
$-2.^9C_2.^7C_3$
0%
$^9C_7.^7C_4$
0%
none of these

The coefficient of

$x^4$ in the expansion of

$(1-2x+3x^2+4x^3)(1-x)^{-8}$ is:

0%
$232$
0%
$231$
0%
$230$
0%
None of these

If the fourth term in the expansion of

$\left(px+\dfrac{1}{x}\right)^{n}$ is independent of

$x$ , then the value of term is :

0%
$5p^{3}$
0%
$10p^{3}$
0%
$20p^{3}$
0%
$None\ of\ these$

The coefficient of

$x^6 y^5 z^3$ in the expansion of

$(3xy - 2xz + zy)^7$ is

0%
$\dfrac {7 !}{4! 2! 1!}\cdot 3^4 \cdot 2^2$
0%
$-\dfrac {7 !}{4! 2! 1!}\cdot 3^4 \cdot 2^2$
0%
$-\dfrac {7 !}{4! 2! 1!}\cdot 3^2 \cdot 2^4$
0%
$\dfrac {7 !}{4! 2! 1!}\cdot 3^2 \cdot 2^4$

Middle term in the expansion of

$(1+3x+3x^2+x^3)^6$ is

0%
$4^{th}$
0%
$3^{rd}$
0%
$10^{th}$
0%
None of these

If the coefficients of

${ x }^{ -2 }$ and

${ x }^{ -4 }$ in the expansion of

${ \left( { x }^{ \frac { 1 }{ 3 } }+\frac { 1 }{ 2x^{ \frac { 1 }{ 3 } } } \right) }^{ 18 },$

$(x>0)$ , are

$m$ and

$n$ respectively, then

$\frac { m }{ n }$ is equal to :

0%
$182$
0%
$\dfrac { 4 }{ 5 }$
0%
$\dfrac { 5 }{ 4 }$
0%
$27$

Explanation

Using binomial theorem,

$\left(x^{\dfrac{1}{3}}+\dfrac{1}{2x^{\dfrac{1}{3}}}\right)^{18}=\displaystyle\sum^{18}_{n=0}{^{18}C_n}\left(x^{\dfrac{1}{3}}\right)^{18-n}\left(\dfrac{1}{2x^{\dfrac{1}{3}}}\right)^n$

$=\displaystyle\sum^{18}_{n=0}{^{18}C_n}x^{\dfrac{18-n}{3}}\cdot \dfrac{1}{2^nx^{n/3}}$

$=\displaystyle\sum^{18}_{n=0}{^{18}C_n}2^{-n}x^{\dfrac{18-n}{3}-\dfrac{n}{3}}$

$=\displaystyle\sum^{18}_{n=0}{^{18}C_n}2^{-n}x^{\dfrac{18-2n}{3}}$

and

$T_{r+1}={^{18}C_r}2^{-r}x^{\dfrac{18-2r}{3}}$

$m=$ coefficient of

$x^{-2}={^{18}C_{12}}2^{-12}=\dfrac{18!}{12!6!}\cdot 2^{-12}$

$\left[\dfrac{18-2r}{3}=-2, 18-2r=-6, \Rightarrow 18+6=2r, \Rightarrow r=\dfrac{24}{2}=12\right]$

$n=$ coefficient of

$x^{-4}={^{18}C_{15}}2^{-15}=\dfrac{18!}{15!3!}2^{-15}$

$\left[18-\dfrac{2r}{3}=-4, \Rightarrow 18-12=-12, \Rightarrow 18+12=2r, \Rightarrow r=\dfrac{30}{2}\right]$

$\therefore \dfrac{m}{n}=\dfrac{\dfrac{18!}{12!6!}2^{-12}}{\dfrac{18!}{15!3!}2^{-15}}$

$=\dfrac{15!3!}{12!6!}2^{12+15}$

$=\dfrac{15\times 14\times 13\times 12!\times 3!}{12!\times 6\times 5\times 4\times 3!}2^3=\dfrac{15\times 14\times 13}{6\times 5\times 4}\times 2^3$

$=\dfrac{91}{4}\times 2^3$

$=182$ .

1196925_1318885_ans_ab644dffae114266a8527c645a8f2926.JPG

The middle term in the expansion of

$\left(\dfrac{2}{3}x-\dfrac{3}{2}y\right)^{20}$ is

0%
$_{ }^{ 20 }{ { C }_{ 10 } }{ x }^{ 10 }{ y }^{ 10 }$
0%
${ x }^{ 10 }{ y }^{ 10 }$
0%
$_{ }^{ 20 }{ { C }_{ 10 } }{ \left( 2/3 \right) }^{ 10 }{ \left( xy \right) }^{ 10 }$
0%
$-x^{ 10 }y^{ 10 }$

Explanation

$\left(\frac{2 }{3} x-\frac{3}{2} y\right)^{20}\\$

$\text { There will be total } 21 \text { terms in this exapnsion } \\$

$\text { : : } 11 \text { th term will be middle tem. }$

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

$\Rightarrow T_{11}={ }^{20} c_{10} \cdot\left(\frac{2}{3} x\right)^{20-10}\left(\frac{-3}{2} y\right)^{10}$

$={}^{20} {c_{10}} x^{10} \cdot y^{10}$

option (a)

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

0%
$2^{n}$
0%
$0$
0%
$3^{n}$
0%
None of these

Explanation

Consider

$(1+x)^n=1+^nC_1x+^nC_2x^2+.........+\,^nC_nx^n$

Put

$x=2$

$3^n=1+2^nC_1+4^nC_2+......+2^n \,^nC_n$

So correct answer is

$3^n$

option (C)

If the constant term in the expansion of

$\left( { x }^{ 3 }-\dfrac { k }{ { x }^{ 8 } } \right)$ is

$1320$ then

$k$ is equal to

0%
$11$
0%
$8$
0%
$2$
0%
$6$

If in the expansion of

$\left(2^{1/3}+\dfrac {1}{3^{1/3}}\right)^{n}$ , the ratio of

$6^{th}$ term from beginning and from the end is

$1/6$ , then the value of

$n$ is

0%
$5$
0%
$7$
0%
$13$
0%
$None\ of\ these$

Explanation

As we know that in an expansion

${\left( a + b \right)}^{n}$ , the general term is given as-

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

In the given expansion-

${\left( {2}^{{1}/{3}} + \cfrac{1}{{3}^{{1}/{3}}} \right)}^{n}$

$a = {2}^{{1}/{3}}, \quad b = \cfrac{1}{{3}^{{1}/{3}}}, \quad n = ?$

Therefore,

$\Rightarrow \cfrac{{^{n}{C}_{5}} {\left( {2}^{{1}/{3}} \right)}^{5} {\left( \cfrac{1}{{3}^{{1}/{3}}} \right)}^{n-5}}{{^{n}{C}_{n-5}} {\left( {2}^{{1}/{3}} \right)}^{n-5} {\left( \cfrac{1}{{3}^{{1}/{3}}} \right)}^{5}} = \cfrac{1}{6} \quad \left( \text{Given} \right)$

$\Rightarrow \cfrac{1}{{\left( {2}^{{1}/{3}} \right)}^{n-10} {\left( {3}^{{1}/{3}} \right)}^{n-10}} = \cfrac{1}{6}$

$\Rightarrow \cfrac{1}{{\left( 6 \right)}^{\frac{n-10}{3}}} = \cfrac{1}{{\left( 6 \right)}^{1}}$

$\Rightarrow \cfrac{n - 10}{3} = 1$

$\Rightarrow n - 10 = 3$

$\Rightarrow n = 10 + 3 = 13$

Find the number of terms in expansion of

$(1+x)^{2}+(1-x)^{8}$

0%
$17$
0%
$18$
0%
$5$
0%
$9$

The term independent of

$x$ in the expansion of

$\left(\sqrt{\dfrac{x}{3}}+\dfrac{3}{2x^{2}}\right)^{10}$ will be

0%
$3$
0%
$5$
0%
$9$
0%
$None\ of\ these$

Explanation

As we know that the general term in an expansion

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

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

Given expansion-

${\left( \sqrt{\cfrac{x}{3}} + \cfrac{3}{2 {x}^{2}} \right)}^{10}$

Here,

$a = \sqrt{\cfrac{}{3}}, \quad b = \cfrac{3}{2 {x}^{2}}, \quad n = 10$

Therefore,

${T}_{r + 1} = {^{10}{C}_{r}} {\left( \sqrt{\cfrac{x}{3}} \right)}^{r} {\left( \cfrac{3}{2 {x}^{2}} \right)}^{10 - r}$

$\Rightarrow {T}_{r + 1} {^{10}{C}_{r}} \; {x}^{\left( \frac{r}{2} - 20 + 2r \right)} {3}^{\left( 10 - r - \frac{r}{2} \right)} {2}^{\left( r - 10 \right)} = {^{10}{C}_{r}} \; {x}^{\left( \frac{5r}{2} - 20 \right)} {3}^{\left( 10 - \frac{3r}{2} \right)} {2}^{\left( r - 10 \right)}$

For the term independent of

$x$ -

$\cfrac{5r}{2} - 20 = 0$

$\Rightarrow r = \cfrac{20 \times 2}{5} = 8$

Hence

${8 + 1}^{th} = {9}^{th}$ term will be independent of

$x$ .

The sum of the coefficients of the first three terms in the expansion of

${\left( {x - \frac{3}{{{x^2}}}} \right)^m},\,x \ne 0$

$m$ being a natural number is ,

$559$ . Find the term of the expansion containing

$x^3$

0%
-5940
0%
1010
0%
1001
0%
1002

Explanation

$\begin{array}{l} { \left( { x-\frac { 3 }{ { { x^{ 2 } } } } } \right) ^{ m } }{ =^{ m } }{ C_{ o } }{ \left( x \right) ^{ m } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ 0 } }{ +^{ m } }{ C_{ 1 } }{ \left( x \right) ^{ m-1 } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ 1 } }{ +^{ m } }{ C_{ 2 } }{ x^{ m-2 } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ 2 } }+...... \\ { =^{ m } }{ C_{ o } }{ \left( x \right) ^{ m } }+{ \left( { -3 } \right) ^{ m } }{ C_{ 1 } }{ \left( x \right) ^{ m-1-2 } }+{ 9.^{ m } }{ C_{ 2 } }{ \left( x \right) ^{ m-2-4 } }+.... \\ { \Rightarrow ^{ m } }{ C_{ o } }+{ \left( { -3 } \right) ^{ m } }{ C_{ 1 } }+{ 9.^{ m } }{ C_{ 2 } }=559 \\ \Rightarrow 1-3m+9\frac { { m\left( { m-1 } \right) } }{ 2 } =559 \\ \Rightarrow 2-6m+9{ m^{ 2 } }-9m=2\times 559=1118 \\ \Rightarrow 9{ m^{ 2 } }-15m-1116=0 \\ \Rightarrow 3{ m^{ 2 } }-15m-372=0 \\ \Rightarrow 3{ m^{ 2 } }-36m+31m-372=0 \\ \Rightarrow 3m\left( { m-12 } \right) +31\left( { m-12 } \right) =0 \\ \Rightarrow \left( { m-12 } \right) \left( { 3m+31 } \right) =0 \\ \therefore m=12,\, \, and\, m=\frac { { -31 } }{ 3 } \left( { does\, not\, exist } \right) \\ so,\, we\, take\, m=12 \\ { \left( { x-\frac { 3 }{ { { x^{ 2 } } } } } \right) ^{ 12 } } \\ Then \\ { T_{ r+1 } }{ =^{ 12 } }{ C_{ r } }{ x^{ 12r } }{ \left( { \frac { { -3 } }{ { { x^{ 2 } } } } } \right) ^{ r } } \\ ={ \left( { -1 } \right) ^{ r } }{ 3^{ r } }^{ 12 }{ C_{ r } }{ x^{ 12-r-25 } } \\ ={ \left( { -1 } \right) ^{ r } }{ 3^{ r } }^{ 12 }{ C_{ r } }{ x^{ 12-3r } } \\ Containg\, { x^{ 3 } } \\ 12-3r=0 \\ \therefore r=3 \\ Now, \\ { T_{ 4 } }={ T_{ 3+1 } }={ \left( { -1 } \right) ^{ 3 } }{ 3^{ 3 } }\, { \, ^{ 12 } }{ C_{ 3 } }{ x^{ 3 } } \\ =-27\, \, { \, ^{ 12 } }{ C_{ 3 } }{ x^{ 3 } } \\ { T^{ 4 } }=-5940{ x^{ 3 } } \end{array}$

hence, the option

$A$ is the correct answer.

If the fourth term in the expansion of

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

$\dfrac { 5 } { 2 } ,$ then

$n + p$ is equal to

0%
$\dfrac { 9 } { 2 }$
0%
$\dfrac { 11 } { 2 }$
0%
$\dfrac { 13 } { 2 }$
0%
$\dfrac { 15 } { 2 }$

The coefficient of

$x^{8}$ in the polynomial

$\left(x-1\right)\left(x-2\right)\left(x-3\right).\left(x-10\right)$ is:

0%
$2640$
0%
$1320$
0%
$1270$
0%
$2740$

The coefficient of

$x ^ { 8 }$ in the expansion of

$\left( 1 + x ^ { 4 } \right) ^ { 3 } ( 1 - x ) ^ { 12 }$ is

0%
$3 + 4 \times 12 C _ { 4 }$
0%
$3 + ^ { 12 } \mathrm { C } _ { 8 }$
0%
$^ { 12 } C _ { 8 }$
0%
$3 - ^ { 12 } \mathrm { C } _ { 8 }$

$C_1 +2C_2 + 3C_3 +4C_4 + ......... + {n+1} nC_n$

0%
$n2^{n-1}$
0%
$2^{n+1}$
0%
$2.2^{n-1}$
0%
Zero

Explanation

$\begin{array}{l} { C_{ 1 } }+2{ C_{ 2 } }+3{ C_{ 3 } }+......+n{ C_{ n } } \\ =n+2\times \frac { { n\left( { n-1 } \right) } }{ { 2! } } +3\times \frac { { n\left( { n-1 } \right) \left( { n-2 } \right) } }{ { 3! } } +....+n\times 1 \\ =n+2\times \frac { { n\left( { n-1 } \right) } }{ 1 } +3\times \frac { { n\left( { n-1 } \right) } }{ 6 } +....+n \\ =n+\frac { { n\left( { n-1 } \right) } }{ 1 } +\frac { { n\left( { n-1 } \right) } }{ 2 } +.....+n \\ =n\left[ { 1+\frac { { n-1 } }{ 1 } +\frac { { n\left( { n-1 } \right) } }{ 2 } +.....+1 } \right] \\ =n\left[ { 1+\frac { { n-1 } }{ { 1! } } +\frac { { n\left( { n-1 } \right) } }{ { 2! } } +.....+1 } \right] \\ put\, \, \, \, \, n-1=N \\ =n\left[ { 1+\frac { N }{ { 1! } } +\frac { { N\left( { N+1 } \right) } }{ { 2! } } +.....+1 } \right] \\ =n\left( { ^{ N }{ C_{ 0 } }{ +^{ N } }{ C_{ 1 } }{ +^{ N } }{ C_{ 2 } }+....{ +^{ N } }{ C_{ N } } } \right) \\ =n{ 2^{ N } } \\ =n{ 2^{ n-1 } } \\ Hence,\, the\, option\, A\, is\, the\, correct\, answer \end{array}$

$\dfrac{C_0}{1} + \dfrac{C_2}{3} + \dfrac{C_4}{5} + \dfrac{C_6}{7} ..... =$

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

Explanation

$\begin{array}{l} \frac { { { C_{ o } } } }{ 1 } +\frac { { { C_{ 2 } } } }{ 3 } +\frac { { { C_{ 4 } } } }{ 5 } +\frac { { { C_{ 6 } } } }{ 7 } ..... \\ 1+\frac { { n\times \left( { n-1 } \right) } }{ { 3! } } +\frac { { n\times \left( { n-1 } \right) \left( { n-2 } \right) \left( { n-3 } \right) } }{ 5 } +...... \\ \frac { 1 }{ { n+1 } } \left[ { \left( { n+1 } \right) \frac { { \left( { n+1 } \right) \times n\times \left( { n-1 } \right) } }{ { 3! } } +\frac { { n\left( { n+1 } \right) \left( { n-1 } \right) \left( { n-2 } \right) } }{ { 5! } } } \right] \\ \frac { 1 }{ { n+1 } } \left[ { { 2^{ n+1-1 } } } \right] \\ \frac { { { 2^{ n } } } }{ { n+1 } } \end{array}$

Hence the option

$A$ is the correct answer.

$(1+x)^{21}+(1+x)^{22}+..+(1+x)^{30}$ in the expansion of this what is the coefficient of

$x^{5}$ is

0%
$^{31} C_5-^{21} C_5$
0%
$^{31} C_4-^{21} C_4$
0%
$^{31} C_6-^{21} C_6$
0%
$None\ of\ these$

The sum

$^ { 20 } \mathrm { C } _ { 0 } + ^ { 20 } \mathrm { C } _ { 1 } + ^ { 20 } \mathrm { C } _ { 2 } + \ldots \ldots . ^ { 20 } \mathrm { C } _ { 10 }$ is equal to

0%
$2 ^ { 19 } + \frac { 20 ! } { ( 10 ! ) ^ { 2 } }$
0%
$2 ^ { 19 } - \frac { 1 } { 2 } \cdot \frac { 20 ! } { ( 10 ! ) ^ { 2 } }$
0%
$2 ^ { 19 } + ^ { 20 } \mathrm { C } _ { 10 }$
0%
none of these

Explanation

As we know that

$(1+x)^{20}=^{20}C_{0}+^{20}C_{1} x+^{20}C_{2} x^2+\cdots+^{20}C_{20} x^{20}$

Put

$x=1$

$\implies 2^{20}=^{20}C_{0}+^{20}C_{1}+^{20}C_2+\cdots+^{20}C_{20}$

$\implies 2^{20}=^{20}C_{0}+^{20}C_{1}+\cdots+^{20}C_{10}+^{20}C_{9}+\cdots+^{20}C_{1}+^{20}C_{0}$

$2^{20}=2(^{20}C_0+^{20}C_1+^{20}C_2+\cdots+^{20}C_{10})-^{20}C_{10}$

$\implies ^{20}C_{0}+^{20}C_1+^{20}C_2+\cdots+^{20}C_{10}=2^{19}+\dfrac{1}{2}^{20}C_{10}=2^{19}+\dfrac{1}{2}\cdot \dfrac{20!}{(10!)^2}$

Coefficient of

$x^ {79}$ in the expansion of

$\left(x+x^ {2}+x^ {4}\right)$ is equal to-

0%
$0$
0%
$150x3^ {19}$
0%
$1$
0%
$3^ {20}$

The number of integral terms in the expansion of

$(\sqrt{3}+\sqrt[8]{5})^{256}$ is

0%
$32$
0%
$33$
0%
$34$
0%
$35$

Explanation

In the expansion of

$(\sqrt3 +\sqrt[8]5)^{256}$ ,

General term,

$T_{r+1}=^{256}C_r (\sqrt 3)^{256-r}(\sqrt [8]{5})^r$

$=\ ^{256}C_r\cdot 3^{(256-r)/2}\cdot 5^{r/8}$

The terms are integral if

$\dfrac{256-r}{2}$ and

$\dfrac{r}{8}$ are both positive integers.

$\therefore r=0, 8, 16, 24,...., 256$

Hence, there are

$33$ integral terns.

For a binomial distribution, n = 5.
If P (X = 4) = P (X = 3), then P (X > 2) is

0%
0.69
0%
0.97
0%
0.21
0%
0.79

Explanation

$n=5\\P(X=4)=^5C_4p^4q=5p^4q\\P(X=3)=^5C_3p^3q^2=10p^3q^2\\P(X=3)=P(X=4)\implies p=2q\\p+q=1\\q=\dfrac 13,p=\dfrac 23\\P(X>2)=P(X=3)+P(X=4)+P(X=5)\\10\left(\dfrac 23\right)^3\left(\dfrac 13\right)^2+5\left(\dfrac 23\right)^4\left(\dfrac 13\right)+\left(\dfrac 23\right)^5=\dfrac{192}{243}=0.79$

The coefficient of

$t^{50}$ in

$(1+t)^{41}(1-t+t^2)^{40}$ is equal to?

0%
$1$
0%
$50$
0%
$81$
0%
$0$

Find the middle term in the expansion of

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

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

Explanation

Given expression

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

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

Middle term

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

$T_{n+1}=^{2n}C_1 (1)^{2n-n} (-x)^n$

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

Middle term is

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

$\left(1+x+x^ {2}+x^ {3}\right)^ {5}=a_{0}+a_{1}x+a_{2}x^ {2}+....+a_{15}x^ {15}$ , then

$a_{10}$ equals

0%
$99$
0%
$100$
0%
$101$
0%
$110$

The largest coefficient in the expansion of

${ \left( 1+x \right) }^{ 38 }$ is

0%
$_{ }^{ 38 }{ { C }_{ 18 } }$
0%
$_{ }^{ 38 }{ { C }_{ 15 } }$
0%
$_{ }^{ 38 }{ { C }_{ 20 } }$
0%
$_{ }^{ 38 }{ { C }_{ 19 } }$

Given that the term of the expansion

$\displaystyle (x^{1/3}+ x^{-1/2})^{15}$ which does not contain

$x$ is

$5$ m , where m

$\in$ N , m

$=$

0%
1100
0%
1010
0%
1001
0%
None of these

Explanation

Given,

$(x^{\dfrac{1}{3}}+x^{-\dfrac{1}{2}})^{15}$

$T_{r+1}=^{15}C_r(x^{\dfrac{1}{3}})^{15-r}(x^{-\dfrac{1}{2}})^r$

Power of

$x=\dfrac{15-r}{3}-\dfrac{r}{2}=0$ for

$x^0$

$2(15-r)-3r=0$

$30-2r-3r=0$

$30-5r=0$

$r=6$

$T_{r+1}=^{15}C_{6}=5005$

$5m=5005\Rightarrow m=1001$

If the

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

$\displaystyle \left [ \dfrac{1}{x^{8/3}} + x^2log_{10}x \right ]^8$ is 5600 , then

$x$ =

0%
10
0%
8
0%
11
0%
9

Explanation

From given, we have,

6th term

$=^8C_5 x^{-\frac{8}{3} \times 3}(x^2 \log x)^5$

$\Rightarrow \dfrac{ 8 \times 7 \times 6}{3 \times 2} \times x^{-8} \times x^{10}(\log x)^5 =5600$

$x^2(\log_{10}x)^5 =100$

$\therefore x=10$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 9 - 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 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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