If $${ \left( 1+2x+x^{ 2 } \right) }^{ n }=\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } } $$, then $${ a }_{ r }=$$

$${ \left( _{ }^{ n }{ { C }_{ r }^{ } } \right) }^{ 2 }$$

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

MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 3

Find the sum

$\displaystyle\sum _{ r=1 }^{ n }{ \cfrac { { r }^{ n }{ C }_{ r } }{ { C }_{ r-1 } } }$

0%
$\displaystyle\cfrac { n(n+1) }{ 3! }$
0%
$\displaystyle\cfrac { n(n-1) }{ 2 }$
0%
$\displaystyle\cfrac { n(n+1) }{ 2 }$
0%
None of these

Explanation

$\displaystyle \sum _{ r=1 }^{ n }{ \frac { { r }^{ n }{ C }_{ r } }{ ^{ n }{ C }_{ r-1 } } } =\sum _{ r=1 }^{ n }{ r\frac { n! }{ r!\left( n-r \right) ! } \times \frac { \left( n-\left( r-1 \right) \right) !\left( r-1 \right) ! }{ n! } }$

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

If the middle term in the expansion of

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

$924\,{ x }^{ 6 }$ , then find the value of

$n$

0%
$n=11$
0%
$n=13$
0%
$n=12$
0%
$n=14$

Explanation

$T_{r+1}=\:^{n}C_{r}x^{2n-3r}$
Therefore power of

$x$ in the middle term is

$6$ .
Hence,

$2n-3r=6$

$\implies r=\dfrac{2n}{3}-2$
The coefficient is

$924$ .
Therefore,

$\:^{n}C_{\frac{2n}{3}-2}=924$

$\implies \dfrac{n!}{\left(\dfrac{n}{3}+2\right)!\left(\dfrac{2n}{3}-2\right)!}=924$
If we substitute

$n=12$ (since it is divisible by 3),
We get

$\:^{n}C_{\frac{2n}{3}-2}$

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

$=924$
Hence

$n=12$

The middle term in the expansion of

$\left (\dfrac{x}{y}+\dfrac{y}{x} \right )^{8}$ is.

0%
$^{8}\textrm{C}_{5}$
0%
$^{8}\textrm{C}_{6}$
0%
$^{8}\textrm{C}_{4}$
0%
$^{8}\textrm{C}_{2}$

Explanation

The middle term of the expression

${ (\dfrac { x }{ y } +\dfrac { y }{ x } ) }^{ 8 }$

There are

$9$ terms in the above expansion.

Hence the middle term will be the

${ 5 }^{ th }$ term

${ T }_{ 5 }={ C }_{ 4 }^{ 8 }{ (\cfrac { x }{ y } ) }^{ 4 }{ (\dfrac { y }{ x } ) }^{ 4 }=$

$^{8}\textrm{C}_{4}$

If the coefficient of (2r + 4)th term is equal to the coefficient of (r - 2)th term in the expansion of

$(1+x)^{18}$ then r

$=$

0%
2
0%
4
0%
6
0%
8

Explanation

Coefficient of (2r + 4)th term

$=^{18}C_{2r+3}$
Coefficient of (r - 2)th term

$=^{18}C_{r-3}$
Coefficient of (2r + 4)th term

$=$ coefficient of (r - 2)th term

$\Rightarrow ^{18}C_{2r+3}=^{18}C_{r-3}$

$\Rightarrow 2r+3+r-3=18$

$\Rightarrow 3r=18 \Rightarrow r=6$

If the coeffecient of the middle term in the expansion of

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

$\alpha$ and coeffecient of middle terms in the expansion of

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

$\beta$ and

$\gamma$ , then relate

$\alpha, \beta$ and

$\gamma$

0%
$\beta -\gamma =\alpha$
0%
$\gamma -\beta =\alpha$
0%
$\beta +\gamma =\alpha$
0%
none of these

Explanation

$(1+x)^{2n+2}$ .
The middle term will be

$(\frac{N}{2}+1)$ th term.

$=\frac{2n+2}{2}+1$ th term

$=(n+2)^{th}$ term
Hence coefficient of the middle term will be

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

$=\alpha$ .
For

$(1+x)^{2n+1}$ .
The middle terms will be

$(\frac{N+1}{2}+1)$ and

$(\frac{N+1}{2})$ terms

$=\frac{2n+2}{2}+1$ th term and

$\frac{2n+2}{2}^{th}$ term.

$=(n+2)^{th}$ term and

$(n+1)^{th}$ term.
Hence coefficient of the middle terms will be

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

$\:^{2n+1}C_{n}=\gamma$
By the properties of binomial coefficients.

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

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

$\beta+\gamma=\alpha$

If the coefficients of

$x^7$ and

$x^8$ in

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

$=$

0%
45
0%
55
0%
35
0%
27

Explanation

Coefficient of $$x^7 = coefficient of $$x^8

$\Rightarrow \displaystyle \frac{2^n \cdot ^nC_7}{6^7} = 2^n \cdot \frac{^n C_8}{6^8}$

$\Rightarrow 6 (^nC_7)=^nC_8$

$\Rightarrow 6 \cdot \displaystyle \frac{n!}{(n-7)!7!}=\frac{n!}{(n-8!)8!}$

$\Rightarrow \displaystyle \frac{6}{n-7}=\frac{1}{8} \Rightarrow n-7 =48$

$\Rightarrow n = 55$

The middle term in the expansion of

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

0%
$\displaystyle\frac { { 2 }^{ n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n! }$
0%
$\displaystyle\frac { { 2n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n! }$
0%
$\displaystyle\frac { { 2n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n-1 }$
0%
$\displaystyle\frac { { 2 }^{ n }{ x }^{ n }(1\cdot 3\cdot 5\cdot \cdot \cdot (2n-1)) }{ n-1 }$

Explanation

The middle term will be

$(\dfrac{N}{2}+1)th$ term.

Hence the middle term here will be

$(\dfrac{2n}{2}+1)th$ term.

$=(n+1)th$ term.

Hence

$r=n$ .

Therefore the coefficient will be

$\:^{2n}C_{n}$

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

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

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

Hence, option A is correct.

In the second term in the expansion

$\displaystyle { \left( \sqrt [ 13 ]{ a } +\frac { a }{ \sqrt { { a }^{ -1 } } } \right) }^{ n }$ is

$\displaystyle 14{ a }^{ \frac { 5 }{ 2 } }$ , then the value of

$\displaystyle \frac { _{ }^{ n }{ { C }_{ 3 }^{ } } }{ ^{ n }{ { C }_{ 2 }^{ } } }$ is

0%
$8$
0%
$12$
0%
$4$
0%
None of these

Explanation

Given:

${ T }_{ 2 }=14{ a }^{\tfrac 52 }$

$\displaystyle \Rightarrow { _{ }^{ n }{ C } }_{ 1 }^{ }{ \left( { a }^{\tfrac 1{13} } \right) }^{ n-1 }.{ \left( \dfrac { a }{ { a }^{ -\tfrac 12 } } \right) }^{ 1 }=14{ a }^{\tfrac 52 }$

$\displaystyle \Rightarrow n.{ a }^{ \dfrac { n-1 }{ 13 } }.{ a }^{ \dfrac { 3 }{ 2 } }=14{ a }^{ \dfrac { 5 }{ 2 } }$

$\displaystyle \Rightarrow n.{ a }^{ \dfrac { n-1 }{ 13 } }=14a\Rightarrow n.{ a }^{ \dfrac { n-14 }{ 13 } }=14\Rightarrow n=14$

$\displaystyle \therefore \dfrac { { _{ }^{ n }{ C } }_{ 3 }^{ } }{ { _{ }^{ n }{ C } }_{ 2 }^{ } } =\dfrac { { _{ }^{ 14 }{ C } }_{ 3 }^{ } }{ { _{ }^{ 14 }{ C } }_{ 2 }^{ } } =4$

The coeffecients of the middle term in the binomial expansion in powers of

$x$ of

${ (1+\alpha x) }^{ 4 }$ and

${ (1+\alpha x) }^{ 6 }$ is the same if

$\alpha$ equals

0%
$-\cfrac { 5 }{ 3 }$
0%
$\cfrac { 10 }{ 3 }$
0%
$\cfrac { 3 }{ 10 }$
0%
$\cfrac { 3 }{ 5 }$

Explanation

The middle terms of the above given binomial expressions will be

$\:^{4}C_{2}(\alpha)^{2}x^2$ and

$\:^{6}C_{3}(\alpha)^{3}x^3$
Since the coefficients are equal,

$\:^{4}C_{2}(\alpha)^{2}=\:^{6}C_{3}(\alpha)^{3}$

$6(\alpha)^{2}=20(\alpha)^{3}$

$\dfrac{3}{10}=\alpha$

The middle term in the expansion of

$({ { x }/{ 2+2) } }^{ 8 }$ is 1120, then

$x\varepsilon R$ is equal to

0%
$-2$
0%
$3$
0%
$-3$
0%
$2$

Explanation

The middle term will be the 5th term.
Hence,

$T_{4+1}=\:^{8}C_{4}(\frac{x}{2})^{4}2^{4}$

$=70x^{4}$

$=1120$ ...(given)
Hence,

$x^4=16$

$x=\pm2$

${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },...,{ C }_{ n }$ are the coefficients of the expansion of

${ \left( 1+x \right) }^{ n }$ , then the value of

$\displaystyle \sum _{ 0 }^{ n }{ \frac { { C }_{ k } }{ k+1 } }$ is

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

Explanation

Here

$\displaystyle { t }_{ r+1 }=\frac { ^{ n }{ { C }_{ r } } }{ r+1 } =\frac { 1 }{ r+1 } .^{ n }{ { C }_{ r } }=\frac { 1 }{ n+1 } .^{ n+1 }{ { C }_{ r+1 } }$

Putting

$r=0,1,2,,...n$ and adding we get,

$\displaystyle \sum _{ 0 }^{ n }{ \frac { ^{ n }{ { C }_{ k } } }{ k+1 } } =\frac { 1 }{ n+1 } \left\{ ^{ n+1 }{ { C }_{ 1 } }+^{ n+1 }{ { C }_{ 2 } }+^{ n+1 }{ { C }_{ 3 } }+...+^{ n+1 }{ { C }_{ n+1 } } \right\}$

$\displaystyle =\frac { 1 }{ n+1 } \left\{ { 2 }^{ n+1 }-^{ n+1 }{ { C }_{ 0 } } \right\} =\frac { { 2 }^{ n+1 }-1 }{ n+1 }$

The middle term in the expansion of

$\displaystyle \left ( 1-\frac{1}{x} \right )^{n}\left ( 1-x \right )^{n},$ is

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

Explanation

$(1-\displaystyle \frac{1}{x})^{n}(1-x)^{n}$

$(\displaystyle \frac{x-1}{x})^{n}(1-x)^{n}$
Case I -

$n$ is even.
Hence the above expression reduces to

$\displaystyle \frac{1}{x^n}(1-x)^{2n}$
The middle term will be

$(n+1)^{th}$ term.
Hence the coefficient will be.

$T_{n+1}=\:^{2n}C_{n}$
Case II -

$n$ is odd.
Then the above expression reduces to.

$-\displaystyle \frac{1}{x^{n}}(1-x)^{2n}$
Middle term will be

$(n+1)^{th}$ term.
Hence coefficient will be

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

If A is the coefficient of the middle term in the expansion of

$\displaystyle (1+x)^{2n}$ and B and C are the coefficients of two middle terms in the expansion of

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

0%
$\displaystyle A+B=C$
0%
$\displaystyle B+C=A$
0%
$\displaystyle C+A=B$
0%
$\displaystyle A+B+C=0$

Explanation

For

$(1+x)^{2n}$ :
The middle term will be

$(\frac{N}{2}+1)$ th term.

$=\frac{2n}{2}+1$ th term

$=(n+1)^{th}$ term
Hence coefficient of the middle term will be

$\:^{2n}C_{n}$

$=A$ .
For

$(1+x)^{2n-1}$ ,
The middle terms will be

$(\frac{N+1}{2}+1)$ and

$(\frac{N+1}{2})$ terms

$=(\frac{2n}{2}+1)^{th}$ term and

$\frac{2n}{2}^{th}$ term.

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

$(n)^{th}$ term.
Hence coefficient of the middle terms will be

$\:^{2n-1}C_{n}=B$ and

$\:^{2n-1}C_{n-1}=C$
By the properties of binomial coefficients.

$\:^{2n-1}C_{n-1}+\:^{2n-1}C_{n}$

$=\:^{2n}C_{n}$
Hence

$B+C=A$

$n$ is an integer between

$0$ and

$21$ , then the minimum value of

$n!(21 - n)!$ is attained for

$n=$

0%
$1$
0%
$10$
0%
$12$
0%
$20$

Explanation

Let us consider

$(1+x)^{21}$
Hence

$T_{r+1}=\:^{21}C_{r}x^{r}$
We know that both the middle terms will have the highest value of binomial coefficient.
The middle terms in thew above binomial expansion are

$11^{th}$ and

$12^{th}$ terms.

$T_{10+1}=\:^{21}C_{10}x^{10}=\dfrac{21!}{10!(11)!}x^{10}$

$T_{11+1}=\:^{21}C_{11}x^{11}=\dfrac{21!}{11!(10)!}x^{11}$

Hence

$n!(21-n)!$ is least for

$n=0, n=21$ followed by

$n=11$ and

$n=10$

The middle term in the expansion of

$\displaystyle (1+x)^{2n}$ is

0%
$\displaystyle \frac{1\cdot 3\cdot5...(2n-1)}{n!}2^{n}\cdot x^{n}$
0%
$\displaystyle \frac{1\cdot 3\cdot5...(2n-1)}{n!}2^{n}\cdot x^{n}x^{n}$
0%
$\displaystyle ^{2n}C_{n}$
0%
$\displaystyle ^{2n}C_{n}-1x^{n-1}$

Explanation

The middle term in the above binomial expansion will be

$(n+1)^{th}$ term.
Therefore coefficient will be

$\:^{2n}C_{n}$

$=\dfrac{2n!}{(n!)(n!)}$

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

If the coefficient of

$x^7$ in

$\left[ax + \left(\displaystyle\frac{1}{bx}\right)\right]^{11} is\ 55 a^{11}$ , then

$a$ and

$b$ satisfy the relation

0%
$a + b = 1$
0%
$a - b = 1$
0%
$ab = 1$
0%
$\displaystyle\frac{a}{b} = 1$

Explanation

General term of

$\left[ax + \left(\displaystyle\frac{1}{bx}\right)\right]^{11}$ is,

$\displaystyle\quad T_{r+1} = ^{11}C_r (ax)^r\cdot \left(\frac{1}{bx} \right)^{11-r} = ^{11}C_ra^rb^{r-11}x^{2r-11}$
Let

$T_{r+1}$ posses

$x^7\Rightarrow 2r-11=7=> r=9$
Hence coefficient of

$x^7$ is

$=^{11}C_9a^9b^{-2}=55a^9b^{-2} = 55a^{11}$ (given)

$\therefore a^2b^2=1\Rightarrow ab = \pm 1$

The middle term in the expansion of

$\displaystyle \left ( x+\frac{1}{x} \right )^{10}$ ,is

0%
$\displaystyle ^{10}C_{1}\frac{1}{x}$
0%
$\displaystyle ^{10}C_{5}$
0%
$\displaystyle ^{10}C_{6}$
0%
$\displaystyle ^{10}C_{7}x$

Explanation

The middle term would be the 6th term.
Hence

$T_{5+1}=\:^{10}C_{5}.$

The value of

$\displaystyle \frac { { _{ }^{ n }{ C } }_{ 1 }^{ } }{ 2 } +\frac { { _{ }^{ n }{ C } }_{ 3 }^{ } }{ 4 } +\frac { { _{ }^{ n }{ C } }_{ 5 }^{ } }{ 6 } +...$ is

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

Explanation

The

$rth$ term in the given expression is

$\displaystyle { T }_{ r }=\frac { { _{ }^{ n }{ C } }_{ 2r-1 }^{ } }{ 2r }$

Since

$\displaystyle \frac { 1 }{ r+1 } .{ _{ }^{ n }{ C } }_{ r }=\frac { 1 }{ n+1 } .{ _{ }^{ n+1 }{ C } }_{ r+1 }$

$\displaystyle \therefore \quad { T }_{ r }=\frac { { _{ }^{ n }{ C } }_{ 2r-1 } }{ 2r } =\frac { 1 }{ n+1 } .{ _{ }^{ n+1 }{ C } }_{ 2r }$

$\displaystyle \therefore \quad \frac { { _{ }^{ n }{ C } }_{ 1 } }{ 2 } +\frac { { _{ }^{ n }{ C } }_{ 3 } }{ 4 } +\frac { { _{ }^{ n }{ C } }_{ 5 } }{ 6 } +....=\frac { 1 }{ n+1 } \left[ { _{ }^{ n+1 }{ C } }_{ 2 }+{ _{ }^{ n+1 }{ C } }_{ 4 }+... \right]$

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

Find the sum 1

$\times$ 2

$\times$ C

$_1$ + 2

$\times$ 3 C

$_2$ + n (n+1)C

$_{n'}$ where C

$_r$ =

$^n$ C

$_r$ .

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

Explanation

Let

$n=1$
Hence

$(n)(n+1)C_{n}$

$=(1)(2)C_{1}$

$=2$ .
This is satisfied by Option A and Option B
Let

$n=2$
Hence

$2\:^{2}C_{1}+(2)(3)\:^{2}C_{2}$

$=4+6$

$=10$
This is satisfied only by Option B
Hence answer is Option B.

$(1+2x+x^2)^n = \displaystyle \sum_{r=0}^{2n} a_r x^r$ , then

$a_r=$

0%
$(^nC_r)^2$
0%
$^nC_r\cdot^nC_{r+1}$
0%
$^{2n}C_r$
0%
$^{2n}C_{r+1}$

Explanation

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

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

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

$T_{r+1}=\:^{2n}C_{r}x^{r}$
Hence

$a_{r}=\:^{2n}C_{r}$

Find the ratio of the coefficient of

${ x }^{ 10 }$ in

${ \left( 1-{ x }^{ 2 } \right) }^{ 10 }$ and the term independent of

$x$ in the expansion of

${ \left( x-\cfrac { 2 }{ x } \right) }^{ 10 }$

0%
$1:8$
0%
$1:16$
0%
$1:24$
0%
$1:32$

Explanation

The general term

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

$(a+b)^{n}$ is

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

In expansion of

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

${T}_{r+1}=\quad^{10}{C}_{r}(1)^{10-r}(-x^{2})^{r}$

$\Rightarrow {T}_{r+1}=\quad^{10}{C}_{r}(-1)^{r}x^{2r}$

$\therefore x^{2r}= x^{10}$

$\Rightarrow 2r=10$

$\Rightarrow r=5$

$coeff({T}_{5+1})={}^{10}{C}_{5}(-1)^{5}= {}^{-10}{C}_{5}$ (equation

$1$ )

In expansion of

$\left(x-\cfrac{2}{x} \right)^{10}$

${T}_{r+1}= \quad^{10}{C}_{r}(x)^{10-r}(\cfrac{-2}{x})^{r}$

$\Rightarrow {T}_{r+1}= \quad^{10}{C}_{r} x^{10-2r}(-2)^{r}$

$\therefore x^{10-2r}=x^{0}$

$\Rightarrow r=5$

$coeff({T}_{5+1}) = {}^{10}{C}_{5}(-2)^{5}= {}^{-10}{C}_{5}(32)$ (equation

$2$ )

By dividing equation

$(1)$ and

$(2)$ we get,

$=\cfrac{^{-10}{C}_{5}}{^{-10}{C}_{5}32}= \cfrac{1}{32}$

Find the coefficient of

$\cfrac { 1 }{ { y }^{ 2 } }$ in

${ \left( y+\cfrac { { c }}{ { y }^{ 2 } } \right) }^{ 10 }$ .

0%
$210{ c }^{ 4 }$
0%
$210{ c }^{ 5 }$
0%
$120{ c }^{ 3 }$
0%
$120{ c }^{ 4 }$

Explanation

$(y+\frac{c}{y^2})^{10}$

$T_{r+1}=\:^{10}C_{r}y^{10-3r}c^r$
Hence for

$y^{-2}$

$10-3r=-2$

$12=3r$

$r=4$
Coefficient will be

$\:^{10}C_{4}c^4$

$=210c^4$

The Value of

$^nC_1+^{n+1}C_2+^{n+2}C_3+...+^{n+m-1}C_m$ is equal

0%
$^{m+n}C_{n-1}$
0%
$^{m+n}C_{n+1}$
0%
$^mC_1+^{m+1}C_2+^{m+2}C_3+...+^{n+m-1}C_n$
0%
$^{m+n}C_{m}-{1}$

Explanation

Adding and subtracting

$\:^{n}C_{0}$ we get

$[\:^{n}C_{0}+\:^{n}C_{1}+\:^{n+1}C_{2}+\:^{n+2}C_{3}...+\:^{n+m-1}C_{m}]-\:^{n}C_{0}$
Applying the formula

$\:^{n}C_{r}+\:^{n}C_{r+1}=\:^{n+1}C_{r+1}$
we get

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

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

$=\:^{n+m-1}C_{m-1}+\:^{n+m-1}C_{m}-1$

$=\:^{n+m}C_{m}-1$

In the expansion of

$(7^{\frac 13} + 11^{\frac 19})^{6561}$ ,

0%
there are exactly $730$ rational terms
0%
there are exactly $5832$ irrational terms
0%
the term which involves greatest binomial coefficient is irrational
0%
the term which involves greatest binomial coefficients is rational

Explanation

Total number of rational terms will be

$1+\dfrac{6561}{L.C.M(3,9)}$

$=1+\dfrac{6561}{9}$

$=730$
Hence total number of irrational terms will be
Total number of terms-total number of rational terms

$=6562-730$

$=5832$
The middle terms are

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

$\left(\dfrac{N+1}{2}+1\right)^{th}$ term.
Both are irrational.

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

$\displaystyle 2{ C }_{ 0 }+{ 2 }^{ 2 }.\frac { { C }_{ 1 } }{ 2 } +{ 2 }^{ 3 }.\frac { { C }_{ 2 } }{ 3 } +...+{ 2 }^{ n+1 }.\frac { { C }_{ n } }{ n+1 } =$

0%
$\displaystyle \frac { { 3 }^{ n+1 }-1 }{ n+1 }$
0%
$\displaystyle \frac { { 3 }^{ n }-1 }{ n }$
0%
$\displaystyle \frac { { 3 }^{ n+2 }-1 }{ n+2 }$
0%
None of these

Explanation

We have,

$\displaystyle { t }_{ r+1 }={ 2 }^{ r+1 }\frac { ^{ n }{ { C }_{ r } } }{ r+1 } ={ 2 }^{ r+1 }\frac { 1 }{ n+1 } .^{ n+1 }{ { C }_{ r+1 } }$

Putting

$r=0,1,2,...,n$ and adding, we get the required sum

$\displaystyle =\frac { 1 }{ n+1 } \left\{ 2.^{ n+1 }{ { C }_{ 1 } }+{ 2 }^{ 2 }.^{ n+1 }{ { C }_{ 2 } }+...+{ 2 }^{ n+1 }.^{ n+1 }{ { C }_{ r+1 } } \right\}$

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

Determine the value of

$x$ in the expression of

${ ( 2+x) }^{ 5 }$ , if the second term in the expansion is

$240$

0%
$(24)^\frac{1}{4}$
0%
$6$
0%
$3$
0%
None of the above

Explanation

The general term of

$(a+x)^n$ is

$^nC_r a^{(n-r)}x^r$

Here,

$n=5$

we will get second term if we put

$r=1$

The second term will be

$^5C_1(2^{(5-1)})x$

$=5(2^{4})x$

$5(2^{4})x=240$ ...... given

Therefore

$5(16)x=240$

$80(x)=240$

$x=3$

Find the

$(n+1)^{th}$ term from the end in the expansion of

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

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

Explanation

There are total

$2n+1$ terms in the given expansion.

Hence

$(n+1)^{th}$ term, from the end is the

$(n+1)^{th}$ term form beginning the beginning also since, it is the middle term.

$T_{n+1}$

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

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

Find the middle term in the expansion of

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

0%
$210$
0%
$630$
0%
$252$
0%
$756$

Explanation

The middle term will be the 6th term. It will also be the only term independent of

$x$ .
Hence the coefficient will be

$T_{5+1}=\:^{10}C_{5}$

$=\displaystyle\frac{10!}{5!(5!)}$

$=252$

The fourth term in the expansion of

${ \left( px+\cfrac { 1 }{ x } \right) }^{ n }$ is

$\cfrac { 5 }{ 2 }$ . Then,

0%
$n=6$
0%
$n=7$
0%
$p=\dfrac{1}{2}$
0%
$p =\dfrac{1}{4}$

Explanation

In the above question

$T_{r+1}=\:^nC_{r}x^{n-2r}p^{n-r}$

$T_{3+1}=\:^nC_{3}x^{n-6}p^{n-3}$

$=\dfrac{5}{2}$
Since it is independent of x,

$n-6=0$

$n=6$
Hence,

$\:^6C_{3}p^{3}=\dfrac{5}{2}$

$20p^3=\dfrac{5}{2}$

$p^3=\dfrac{1}{8}$

$p=\dfrac{1}{2}$

${ \left( 1+2x+x^{ 2 } \right) }^{ n }=\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } }$ , then

${ a }_{ r }=$

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

Explanation

We have

$\displaystyle { \left( 1+2x+{ x }^{ 2 } \right) }^{ n }=\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } }$

$\displaystyle \Rightarrow { \left( 1+x \right) }^{ 2n }=\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } }$

$\displaystyle \Rightarrow \sum _{ r=0 }^{ 2n }{ ^{ 2n }{ { C }_{ r } }{ x }^{ r } } =\sum _{ r=0 }^{ 2n }{ { a }_{ r }{ x }^{ r } } \Rightarrow { a }_{ r }=^{ 2n }{ { C }_{ r } }$

The middle term in the expansion of

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

$n$ being a positive integer is

0%
$\cfrac { \left\{ 1.3.5....(2n) \right\} { 2 }^{ n } }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad$
0%
$\cfrac { \left\{1.3.5....(2n) \right\} { 2 }^{ n }n! }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad$
0%
$\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n }n! }{ n! } { x }^{ n }\\ \quad \quad \quad \quad \quad \quad \quad \quad$
0%
$\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n } }{ n! } { x^n }\\ \quad \quad \quad \quad \quad \quad \quad \quad$

Explanation

The number of terms in the expansion of

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

$(2n+1)$ (odd), its middle term is

$\dfrac{2n+1+1}2=(n+1)^{th}$ term.

$\therefore$ Required term

$={ T }_{ n+1 }={ _{ }^{ 2n }{ C } }_{ n }{ x }^{ n }\\ =\cfrac { 2n! }{ n!n! } { x }^{ n }=\cfrac { \left\{ 1.2.3.4.5.6...(2n-1)2n \right\} }{ n!n! } { x }^{ n }\\ =\cfrac { \left\{ 1.3.5....(2n-1) \right\} \left\{ 2.4.6.....2 \right\} n }{ n!n! } { x }^{ n }\\ =\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n }(1.2.3....n) }{ n!n! } { x }^{ n }\\ =\cfrac { \left\{ 1.3.5....(2n-1) \right\} { 2 }^{ n } }{ n! } { x }^{ n }\\$

$n>2$ , then find the value of

${ C }_{ 1 }{ \left( a-1 \right) }^{ 2 }-{ C }_{ 2 }{ \left( a-2 \right) }^{ 2 }+{ C }_{ 3 }{ \left( a-3 \right) }^{ 2 }-.....+{ \left( -1 \right) }^{ n-1 }{ C }_{ n }{ \left( a-n \right) }^{ 2 }$ where

${ C }_{ r }$ stands for

$\quad { _{ }^{ n }{ C } }_{ r }$

0%
${ a }^{ 3}$
0%
${ a }$
0%
$\dfrac{ a }{2}$
0%
${ a }^{ 2 }$

Explanation

Let

$n=3$
Hence the above expression reduces to

$\:^{3}C_{1}(a-1)^{2}-\:^{3}C_{2}(a-2)^{2}+\:^{3}C_{3}(a-3)^{3}$

$=3(a-1)^{2}-3(a-2)^{2}+(a-3)^{3}$

$=3a^2-6a+3-(3a^2-12a+12)+(a^2-6a+9)$

$=a^2$
Hence answer is Option D

find the 7th term in the expansion of

${ \left( 4x-\frac { 1 }{ 2\sqrt { x } } \right) }^{ 13 }$

0%
$439296{ x }^{ 7 }$
0%
$439296{ x }^{ 4 }$
0%
$439396{ x }^{ 7 }$
0%
$43396{ x }^{ 4 }$

Explanation

The 7th term of the given binomial expansion will be,

$=$

$\displaystyle _{6}^{13}\textrm{C} \; (4x)^{7} \;(-\frac{1}{2x^{\frac{1}{2}}})^{6}$

$=$

$\displaystyle\frac{13! \times 2^{14} \times x^{7}}{7! . 6! \times 2^{6}.x^{3}}$

$=$

$\displaystyle\frac{13! \times 2^{14}}{7! \times 6! \times 2^{6}}x^{4}$

$=439296x^{4}$

The

$4th$ term from the end in the expansion of

${ \left( \cfrac { { x }^{ 3 } }{ 2 } -\cfrac { 2 }{ { x }^{ 2 } } \right) }^{ 7 }$ is

0%
$35x$
0%
$70x^{2}$
0%
$35x^{2}$
0%
$70x$

Explanation

For the above question

$T_{r+1}=\:^{7}C_{r}x^{21-5r}2^{2r-7}$
For the fourth term, from the end

$r=4$

$T_{5+1}=\:^{7}C_{4}x^{1}2^{}$

$=(35)(2)x$

$=70x$

The middle term in the expansion of

${ \left( \cfrac { a }{ x } +bx \right) }^{ 12 }$ is

0%
$924{ a }^{ 6 }{ b }^{ 6 }$
0%
$924{ a }^{ 6 }{ b }^{ 5 }$
0%
$924{ a }^{ 5 }{ b }^{ 5 }$
0%
$924{ a }^{ 5 }{ b }^{ 6 }$

Explanation

The middle term will be the 7th term.
Hence

$T_{6+1}=\:^{12}C_{6}(\frac{a}{x})^{6}(bx)^{6}=924a^{6}b^{6}$

The

$8^{th}$ term of

$\displaystyle { \left( 3x+\frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 12 }$ , when expanded ina scending power of

$x$ , is

0%
$\displaystyle \frac { 228096 }{ { x }^{ 3 } }$
0%
$\displaystyle \frac { 228096 }{ { x }^{ 9 } }$
0%
$\displaystyle \frac { 328179 }{ { x }^{ 3 } }$
0%
None of these

Explanation

When

$\displaystyle { \left( 3x+\frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 12 }$ is expanded, the power of

$x$ goes on decreasing as the terms proceed.

hence, it is expanded in descending power of

$x$ .

$\displaystyle { \left( \frac { 2 }{ 3{ x }^{ 2 } } +3x \right) }^{ 12 }$ , when expanded will be ascending power of

$x$ .

Now

${ t }_{ 8 }$ in

$\displaystyle { \left( \frac { 2 }{ 3{ x }^{ 2 } } +3x \right) }^{ 12 }$

$\displaystyle =_{ }^{ 12 }{ { C }_{ 7 }^{ } }{ \left( \frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 12-7 }.{ \left( 3x \right) }^{ 7 }$

$\displaystyle =\frac { 12! }{ 7!5! } .{ \left( \frac { 2 }{ 3{ x }^{ 2 } } \right) }^{ 5 }.{ \left( 3x \right) }^{ 7 }=\frac { 12\times 11\times 10\times 9\times 8 }{ 5\times 4\times 3\times 2 } .\frac { { 2 }^{ 5 }.{ 3 }^{ 2 } }{ { x }^{ 3 } }$

$\displaystyle =\frac { 228096 }{ { x }^{ 3 } }$

Find the middle term in the expansion of

${ \left( 3x-\cfrac { { x }^{ 3 } }{ 6 } \right) }^{ 9 }$ .

0%
${_{ }^{ 9 }{ C } }_{ 6 }{ \left( 3x \right) }^{ 5 }$
0%
${ _{ }^{ 9 }{ C } }_{ 5 }{ \left( 3x \right) }^{ 4}$
0%
Both A & B
0%
none of the above

${ \left( 8+3\sqrt { 7 } \right) }^{ n }=\alpha +\beta$ where

$n$ and

$\alpha$ are positive integers and

$\beta$ is a positive proper fraction,then

0%
$(1-\beta)(\alpha+\beta)=1$
0%
$(1+\beta)(\alpha+\beta)=1$
0%
$(1-\beta)(\alpha-\beta)=1$
0%
$(1+\beta)(\alpha-\beta)=1$

Explanation

Given,

$\alpha+\beta=(8+3\sqrt{7})^{n}$

And

$1-\beta$

$=\dfrac{1}{\alpha+\beta}$

$=\dfrac{1}{(8+3\sqrt{7})^{n}}$

Rationalizing, we get

$1-\beta$

$=\dfrac{1}{(8+3\sqrt{7})^{n}}\dfrac{(8-3\sqrt{7})^{n}}{(8-3\sqrt{7})^{n}}$

$=\dfrac{(8-3\sqrt{7})^{n}}{(64-63)^n}$ ........

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

$=(8-3\sqrt{7})^{n}$

Now

$(1-\beta)(\alpha+\beta)$

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

$=(64-63)^{n}$ ........

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

$=1$

Option A is correct

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

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

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

Explanation

$\sum _{ 0\le i\le }^{ }{ \sum _{ j\le n }^{ }{ { \left( { C }_{ i }+{ C }_{ j } \right) }^{ 2 } } } \quad i=0,1,2,...,\left( n-1 \right) ;j=1,2,3,...,n$ and

$i<j$

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

Find the middle term in the expansion of

${ \left( \cfrac { x }{ a } -\cfrac { a }{ x } \right) }^{ 21 }$

0%
${ _{ }^{ 20 }{ C } }_{ 10 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x }$
0%
${ _{ }^{ 20 }{ C } }_{ 9 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x }$
0%
${ _{ }^{ 21 }{ C } }_{ 10 }\cfrac { x }{ a } , -{ {}_{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x }$
0%
${ _{ }^{ 21 }{ C } }_{ 9 }\cfrac { x }{ a } , { _{ }^{ 21 }{ C } }_{ 10 }\cfrac { a }{ x }$

Explanation

There are 22 terms in the expansion of

$(\displaystyle \frac{x}{a}-\displaystyle \frac{a}{x})^{21}$ .
So, the middle terms are

$T_{11}$ and

$T_{12}$

$T_{11}=^{21}\textrm{C}_{10}(\displaystyle \frac{x}{a})^{11}(\displaystyle \frac{-a}{x})^{10}=^{21}\textrm{C}_{10}(\displaystyle \frac{x}{a})$
and

$T_{12}=^{21}\textrm{C}_{11}(\displaystyle \frac{x}{a})^{10}(\displaystyle \frac{-a}{x})^{11}=-^{21}\textrm{C}_{11}(\displaystyle \frac{a}{x})=-^{21}\textrm{C}_{10}(\displaystyle \frac{a}{x})$

If the second term in the expansion

${ \left[ a^{\dfrac {1}{13}} +\dfrac { a }{ \sqrt { { a }^{ -1 } } } \right] }^{ n }$ is

$14\ { a }^{ 5/2 }$ , then the value of

$\dfrac {^{n}C_{3}}{^{n}C_{2}}$ is

0%
$4$
0%
$3$
0%
$12$
0%
$6$

Explanation

The first term will have a coefficient of

$\:^{n}C_{1}=n$
Now, this coefficient has to be equal to

$14.$
Therefore

$n=14$
Now

$\dfrac{\:^{14}C_{3}}{\:^{14}C_{2}}=\dfrac{14!(12!)(2!)}{11!(3!)(14!)}=\dfrac{12}{3}=4$

If the number of terms in

${ \left( x+1+\cfrac { 1 }{ x } \right) }^{ n }\quad (n\in { I }^{ + })$ is 401, then

$n$ is greater than

0%
201
0%
200
0%
199
0%
none of these

Explanation

If we substitute 2 in place of 1 in the above expression, we get

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

$((\sqrt{x}+\frac{1}{\sqrt{x}}))^{2n}$
Since the total number of terms is 401

$2n+1=401$

$2n=400$

$n=200$
However the actual question is

$(x+1+\frac{1}{x})^{n}$
Hence n is greater than 199.

$\displaystyle{ a }_{ n }=\sum _{ r=0 }^{ n }{ \cfrac { 1 }{ { _{ }^{ n }{ C } }_{ r } } }$ then

$\displaystyle\sum _{ r=0 }^{ n }{ \cfrac { r }{ { _{ }^{ n }{ C } }_{ r } } }$ equals

0%
$(n-1){ a }_{ n }$
0%
$n{ a }_{ n }$
0%
$\cfrac { 1 }{ 2 } n{ a }_{ n }$
0%
$\cfrac { (n-1) }{ 2 } { a }_{ n }$

Explanation

We have

$\displaystyle { a }_{ n }=\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ \frac { n }{ 2+1 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } +\frac { 1 }{ ^{ n }{ { C }_{ n-r } } } }$

$\displaystyle \Rightarrow { a }_{ n }=\sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 2 }{ ^{ n }{ C_{ r } } } =2 } \sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } }$

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

$\displaystyle \Rightarrow n\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =n\left( \frac { { a }_{ n } }{ 2 } \right) =\frac { n }{ 2 } { a }_{ n }$

The total number of terms in the expansion of

${ \left( x+a \right) }^{ 100 }+{ \left( x-a \right) }^{ 100 }$ after simplification is

0%
202
0%
51
0%
50
0%
49

Explanation

In the above binomial expansion, the terms at the even places will get eliminated, and we would be left with twice the sum of the terms at odd places.
Hence there will be

$\frac{n}{2}+1$

$=\frac{100}{2}+1$

$=51$ terms

Let

$n$ and

$k$ be positive integers such that

$\displaystyle n\ge \frac { k\left( k+1 \right) }{ 2 } .$ The number of solution

$\left( { x }_{ 1 },{ x }_{ 2 },..,{ x }_{ k } \right) \ge 1;{ x }_{ 2 }\ge 2,...,{ x }_{ k }\ge k$ all integers satisfying

${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...+{ x }_{ k }=n$ is

0%
$^{ m }{ { C }_{ k-1 } }$
0%
$^{ m }{ { C }_{ k } }$ 3
0%
$^{ m }{ { C }_{ k+1 } }$
0%
None of these

Explanation

The number of solutions of

${ x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }+...{ x }_{ k }=n$

$=$ coefficients of

${ t }^{ n }$ in

$\left( t+{ t }^{ 2 }+{ t }^{ 3 }+.. \right) \left( { t }^{ 2 }+{ t }^{ 3 }+.. \right) ...\left( { t }^{ k }+{ t }^{ k+1 }+.. \right)$

$=$ coefficient of

${ t }^{ n }$ in

${ t }^{ 1+2+3+...k }{ \left( 1+t+{ t }^{ 2 }+{ t }^{ 3 }+.. \right) }^{ k }$

But

$1+2+3+...k=k\displaystyle\frac { k+1 }{ 2 } =r$

and

$1+t+{ t }^{ 2 }+{ t }^{ 3 }+..=\displaystyle\frac { 1 }{ 1-t }$

Thus the required number of solutions

$=$ coefficients of

${ t }^{ n-r }$ in

${ \left( 1-t \right) }^{ -k }$

$=$ coefficients of

${ t }^{ n-r }$ in

$1+_{ }^{ k }{ { C }_{ 1 }^{ }t+ }^{ k+1 }{ { C }_{ 2 }^{ }{ t }^{ 2 }+.... }$

$\Rightarrow \ \ _{ }^{ k+n-r-1 }{ { C }_{ n-r }^{ } }=^{ k+n-r-1 }{ { C }_{ k-1 }^{ } }=^{ m }{ { C }_{ k-1 }^{ } }$

The number of irrational terms in the expansion of

${ \left( { 2 }^{ \dfrac 15 }+{ 3 }^{ \dfrac {1}{10} } \right) }^{ 55 }$ is

0%
$47$
0%
$56$
0%
$50$
0%
$48$

Explanation

For the above question

$T_{r+1}=\:^{55}C_{r}2^{11-\dfrac{r}{5}}3^{\dfrac{r}{10}}$
Hence we will have rational terms at

$r=0,10,20,30,40,50$ respectively.
Hence there will be

$6$ rational terms.
The total number of terms will be

$55+1$

$=56$ terms.
Hence the number of irrational terms will be

$56-6$

$=50$ terms.

$\cfrac { { _{ }^{ n }{ C } }_{ r }+4{ _{ }^{ n }{ C } }_{ r+1 }+6{ _{ }^{ n }{ C } }_{ r+2 }+4{ _{ }^{ n }{ C } }_{ r+3 }+{ _{ }^{ n }{ C } }_{ r+4 } }{ { _{ }^{ n }{ C } }_{ r }+3{ _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+{ _{ }^{ n }{ C } }_{ r+3 } } =\cfrac { n+k }{ r+k }$ . Find the value of k

0%
2
0%
4
0%
6
0%
8

Explanation

$\displaystyle\cfrac { { _{ }^{ n }{ C } }_{ r }+4{ _{ }^{ n }{ C } }_{ r+1 }+6{ _{ }^{ n }{ C } }_{ r+2 }+4{ _{ }^{ n }{ C } }_{ r+3 }+{ _{ }^{ n }{ C } }_{ r+4 } }{ { _{ }^{ n }{ C } }_{ r }+3{ _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+{ _{ }^{ n }{ C } }_{ r+3 } }$

$\displaystyle=1+\cfrac { { _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+3{ _{ }^{ n }{ C } }_{ r+3 }+{ _{ }^{ n }{ C } }_{ r+4 } }{ { _{ }^{ n }{ C } }_{ r }+3{ _{ }^{ n }{ C } }_{ r+1 }+3{ _{ }^{ n }{ C } }_{ r+2 }+{ _{ }^{ n }{ C } }_{ r+3 } }$
using

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

$\displaystyle=1+\cfrac { { _{ }^{ n+1 }{ C } }_{ r+2 }+2{ _{ }^{ n+1 }{ C } }_{ r+3 }+{ _{ }^{ n+1 }{ C } }_{ r+4 } }{ { _{ }^{ n+1 }{ C } }_{ r+1 }+2{ _{ }^{ n+1 }{ C } }_{ r+2 }+{ _{ }^{ n+1 }{ C } }_{ r+3 } }$

$\displaystyle=1+\cfrac { { _{ }^{ n+2 }{ C } }_{ r+3 }+{ _{ }^{ n+2 }{ C } }_{ r+4 } }{ { _{ }^{ n+2 }{ C } }_{ r+2 }+{ _{ }^{ n+2 }{ C } }_{ r+3 } }$

$\displaystyle=1+\cfrac { { _{ }^{ n+3 }{ C } }_{ r+4 } }{ { _{ }^{ n+3 }{ C } }_{ r+3 } }$

$=\displaystyle\frac{n+4}{r+4}$

$\therefore k=4$
Hence, option B.

In the expansion of

$\left (x+ \sqrt{x^{2}-1}\right )^{6}$ +

$\left (x- \sqrt{x^{2}-1}\right )^{6}$ ,the number of terms is

0%
$7$
0%
$14$
0%
$6$
0%
$4$

Explanation

Let

$a=x,b=\sqrt{x^2-1}$
Now,

$(x+\sqrt{x^2-1})^6=(a+b)^6$

$=^6C_0a^6+^6C_1a^5b+^6C_2a^4b^2+^6C_3a^3b^3+^6C_4a^2b^4+^6C_5ab^5+^6C_6 b^6$
And

$(x-\sqrt{x^2-1})^6= (a-b)^6$

$=^6C_0a^6-^6C_1a^5b+^6C_2a^4b^2-^6C_3a^3b^3+^6C_4a^2b^4-^6C_5ab^5+^6C_6 b^6$
Now adding above two equations,

$(a+b)^6+(a-b)^6=2(^6C_0a^6+^6C_2a^4b^2+^6C_4a^2b^4+^6C_6b^6)$
Hence number of terms in the required expansion is

$4$

The number of real negative terms in the binomial expansion of

$\left ( 1+ix \right )^{4n-2},$

$n\epsilon N,$

$x>0,$ is

0%
$n$
0%
$n+1$
0%
$n-1$
0%
$2n$

Explanation

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

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

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

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

Total number of terms will be

$2n-1+1=2n$ .

Hence the number of real negative terms will therefore be

$=\dfrac{2n}{2}$

$=n$ .

Find the sum of the series

$\displaystyle\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }\left[ \cfrac { 1 }{ { 2 }^{ r } } +\cfrac { { 3 }^{ r } }{ { 2 }^{ 2r } } +\cfrac { { 7 }^{ r } }{ { 2 }^{ 3r } } +\cfrac { { 15 }^{ r } }{ { 2 }^{ 4r } } +...upto\: m\: terms \right]$

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

Explanation

$\displaystyle \sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } }. { _{ }^{ n }{ C } }_{ r }\left[ \cfrac { 1 }{ { 2 }^{ r } } +\cfrac { { 3 }^{ r } }{ { 2 }^{ 2r } } +\cfrac { { 7 }^{ r } }{ { 2 }^{ 3r } } +\cfrac { { 15 }^{ r } }{ { 2 }^{ 4r } } +... \right]$ upto m terms

$\displaystyle=\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ \left( \frac { 1 }{ 2 } \right) }^{ r }+\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ \left( \frac { 3 }{ 4 } \right) }^{ r }+\sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }\left( \right) { \left( \frac { 7 }{ 8 } \right) }^{ r }+...$

$\displaystyle={ \left( 1-\frac { 1 }{ 2 } \right) }^{ n }+{ \left( 1-\frac { 3 }{ 4 } \right) }^{ n }+{ \left( 1-\frac { 7 }{ 8 } \right) }^{ n }+...\quad \quad \left[ \because \sum _{ r=0 }^{ n }{ { \left( -1 \right) }^{ n } } { _{ }^{ n }{ C } }_{ r }{ x }^{ r }={ \left( 1-x \right) }^{ r } \right]$

$\displaystyle={ \left( \frac { 1 }{ 2 } \right) }^{ n }+{ \left( \frac { 1 }{ 4 } \right) }^{ n }+{ \left( \frac { 1 }{ 8 } \right) }^{ n }+...={ \left( \frac { 1 }{ 2 } \right) }^{ n }\left[ \frac { 1-{ \left( \frac { 1 }{ { 2 }^{ n } } \right) }^{ m } }{ 1-{ \left( \frac { 1 }{ 2 } \right) }^{ n } } \right]$

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

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 3 - 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 3 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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