MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 7

The value of

$C_1 ^2+C_2 ^2....+C_n ^2$ (where

$C_i$ is the

$i^{th}$ coefficient of

$(1+x)^n$ expansion), is:

0%
$\dfrac {n^n}{n!}$
0%
$\dfrac {2n!}{n!n!}$
0%
$\dfrac {2n!}{n!}$
0%
$\dfrac {n!\times2^n}{2n!}$

Explanation

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

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

Now,

$\quad \sum _{ r=0 }^{ n }{ { \left( ^{ n }{ { C }_{ r } } \right) }^{ 2 } }$ = co-efficient of

$x^n$ in

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

Thus,

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

${ \left( ^{ n }{ { C }_{ i } } \right) }$ is the

$i^{th}$ coefficient of

$(1+x)^n$ expansion.

So,

$C_1^2+C_2^2+...+C_n^2={ \left( ^{ 2n }{ { C }_{ n } } \right) }=\dfrac { 2n! }{ n!n! }$

$f(n)=\sum_{s=1}^n \sum_{r=s}^n \:^nC_r \:^rC_s$ , then

$f(3)=$

0%
$27$
0%
$19$
0%
$1$
0%
$5$

Explanation

Given :

$f\left( n \right) =\overset { n }{ \underset { s=1 }{ \Sigma } }\; \overset { n }{ \underset { r=s }{ \Sigma } } { ^{ n }{ C } }_{ r }{ ^{ r }{ C } }_{ s }$

To Find :

$f\left( 3 \right) =?$

Sol. :

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

$+\left\{ { ^{ n }{ C } }_{ n-1 }{ ^{ n-1 }{ C } }_{ n-1 }+{ ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ n-1 } \right\} +\left\{ { ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ n } \right\}$

$\Rightarrow$ Take

$n=3$

$\Rightarrow f\left( 3 \right) =\left\{ { ^{ 3 }{ C } }_{ 1 }{ ^{ 1 }{ C } }_{ 1 }+{ ^{ 3 }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 1 }+{ ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 1 } \right\} +\left\{ { ^{ 3 }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 2 }+{ ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 2 } \right\} +\left\{ { ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 3 } \right\}$

$=\left\{ 3+3.2+3 \right\} +\left\{ 3+3 \right\} +\left\{ 1 \right\}$

$=19$

Hence, the answer is

$19.$

The value of

$(n+2)C_02^{n+1}-(n+1)C_12^n+nC_22^{n-1}+....$ is equal to:

$(C_r=\:^nC_r)$

0%
$4n$
0%
$4$
0%
$2n+4$
0%
$4+4n$

Explanation

We know,

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

Multiplying by

$x^2$ on both sides,

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

Taking the derivative,

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

Putting

$x=2$ in LHS, we get our required result,

$\quad4(-1)^n+n(4)(-1)^{n-1}$

$=4-4n$ ,

$n$ is even

$=4n-4$ ,

$n$ is odd.

There is

$no$

$option$ matching the answer.

$^{n-1}C_{r}=(k^2-3)\:^nC_{r+1}$ , then

$k\:\:\epsilon$

0%
$(-\infty,-2]$
0%
$[2,\infty)$
0%
$[-\sqrt 3,\sqrt 3]$
0%
$(\sqrt 3,2]$

Explanation

Given,

$^{n-1}{C}_{r}=(k^2-3)^{n}{C}_{r+1}$

$\Rightarrow$

$\dfrac{^{n-1}{C}_{r}}{^{n}{C}_{r+1}}=(k^2-3)$ ....Using

$^{n}{C}_{x}=\dfrac{n}{x}.^{n-1}{C}_{x-1}$

$\Rightarrow$

$\dfrac{r+1}{n}=(k^2-3)$ ....As

$n>r+1$

So,

$0<\dfrac{r+1}{n}\le1$

Therefore,

$0<k^2-3\le1$

$\Rightarrow 3<k^2\le4$

$\Rightarrow \sqrt{3}<k\le2$

So, k

$\in$

$(\sqrt{3},2]$

$P_n$ denotes the product of all the coefficients in the expansion of

$(1+x)^n$ , then

$\dfrac {P_{n+1}}{P_n}$ is equal to:

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

Explanation

$(1+x)^n$

${ P }_{ n+1 }=_{ }^{ n+1 }{ { C }_{ 0 } }\times _{ }^{ n+1 }{ { C }_{ 1 } }\times _{ }^{ n+1 }{ { C }_{ 2 } }\times ........\times _{ }^{ n+1 }{ { C }_{ n+1 } }\\ { P }_{ n }=_{ }^{ n }{ { C }_{ 0 } }\times _{ }^{ n }{ { C }_{ 1 } }\times _{ }^{ n }{ { C }_{ 2 } }\times ........\times _{ }^{ n }{ { C }_{ n } }$

$\quad \prod _{ r=0 }^{ n }{ \dfrac { _{ }^{ n+1 }{ { C }_{ r } } }{ _{ }^{ n }{ { C }_{ r } } } \quad \quad \quad \quad \quad \quad \quad \left( \because _{ }^{ n+1 }{ { C }_{ n+1 }=1 } \right) } \\ =\prod _{ r=0 }^{ n }{ \dfrac { (n+1)! }{ r!(n+1-r)! } } \times \dfrac { r!(n-r)! }{ n! } \\ =\prod _{ r=0 }^{ n }{ \dfrac { (n+1) }{ \left( n-r+1 \right) } } \\ ={ \left( n+1 \right) }^{ n }\prod _{ r=0 }^{ n }{ \dfrac { 1 }{ n-r+1 } } \\ =\dfrac { { \left( n+1 \right) }^{ n } }{ \left( n+1 \right) ! }$

Hence, the correct option is

$D$ .

The value of

$\displaystyle \:^{50}C_4+\sum_{r=1}^6 \:^{56-r}C_3$ is

0%
$\:^{55}C_4$
0%
$\:^{55}C_3$
0%
$\:^{56}C_3$
0%
$\:^{56}C_4$

Explanation

$^{50}{C}_{4}$ +

$\sum _{ r=1 }^{6 }{^{56-r}{C}_{3} }$

$^{50}{C}_{4}$ +

$^{50}{C}_{3}$ +

$^{51}{C}_{3}$ +.................+

$^{55}{C}_{3}$ Using

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

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

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

$^{51}{C}_{4}$ +

$^{51}{C}_{3}$ +.................+

$^{55}{C}_{3}$

Similarly the series would become like this

$^{55}{C}_{4}$ +

$^{55}{C}_{3}$ Using

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

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

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

$^{56}{C}_{4}$

The coefficient of

$x^{53}$ in the following expansions.

$\displaystyle \sum_{m=0}^{100} \,^{100}C_m(x-3)^{100-m}\cdot 2^m$ is

0%
$^{100}C _{47}$
0%
$^{100}C _{53}$
0%
$^{-100}C _{53}$
0%
$^{-100}C _{100}$

Explanation

$\displaystyle \sum_{m=0}^{100} \, ^{100}C_m(x-3)^{100-m}\cdot 2^m$

Above expansion can be rewritten as

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

$\therefore x^{53}$ will occur in

$T_{54}$ i.e.

$54^{th}$ term.

So,

$m=53$

$\Rightarrow T_{54}={ }^{100}C_{53}(-x)^{53}$

$\therefore$ Required coefficient is

${ }^{-100} C_{53}$ .

The coefficient of

$x^{2012}$ in

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

0%
1
0%
2
0%
3
0%
4

Explanation

$\displaystyle{\frac { (1+x) }{ (1+{ x }^{ 2 })(x-1) } =\frac { x }{ (1+{ x }^{ 2 }) } +\frac { 1 }{ (1-x) } }$

The general term for

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

$\displaystyle{{ \left( 1-x \right) }^{ -1 }=\sum _{ 0 }^{ \infty }{ { x }^{ n } } .........(i)}$

Replacing

$x$ by

$-{ x }^{ 2 }$

$\displaystyle{\frac { 1 }{ 1+{ x }^{ 2 } } =1-{ x }^{ 2 }+{ x }^{ 4 }-{ x }^{ 6 }.......\\ \frac { 1 }{ 1+{ x }^{ 2 } } =\sum _{ 0 }^{ \infty }{ { { (-1) }^{ n }x }^{ 2n } }}$

Multiplying both sides by

$x$

$\displaystyle{\frac { x }{ 1+{ x }^{ 2 } } =\sum _{ 0 }^{ \infty }{ { { (-1) }^{ n }x }^{ 2n+1 } } ....(ii)}$

From (i) clearly the cofficient of

${ x }^{ 2012 }$ is

$1$ and there is no term with even cofficient in

$(ii)$

So the cofficient of

$\displaystyle{{ x }^{ 2012 }}$ in the given expression is

$1$

Let

$n$ be a positive integer and,

${ \left( 1+x \right) }^{ n }={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+\dots +{ a }_{ n }{ x }^{ n }$ . What is

${ a }_{ 0 }+{ a }_{ 1 }+{ a }_{ 2 }+\dots +{ a }_{ n }$ equal to?

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

Explanation

Given expansion is

${(1+x)}^n=a_0+a_1 x+a_2 x^2+....a_n x^n$

Substituting

$x=1$ on both the sides we get,

${(1+1)}^n=a_0+a_1\times 1+a_2\times 1^2+....a_n\times 1^n$

$\implies 2^n=a_0+a_1+a_2+...a_n$

Thus, the value of

$a_0+a_1+a_2+...a_n$ is

$2^n$ .

$5^{th}$ term from the end in the expansion of

$\left( \dfrac{x^2}{2} - \dfrac{2}{x^2} \right )^{12}$ is

0%
$-7920x^{-4}$
0%
$7920x^{4}$
0%
$7920x^{-4}$
0%
$-7920x^{4}$

Explanation

${ 5 }^{ th }$ term in the expansion of

${ \left( \cfrac { { x }^{ 2 } }{ 2 } -\cfrac { 2 }{ { x }^{ 2 } } \right) }^{ 12 }$ is

${ T }_{ r+1 }= { }^nC_r { x }^{ r }.{ y }^{n- r }$

${ T }_{ 5 }= { }^{12}C_4 \left[{ \left( { \cfrac { { x }^{ 2 } }{ 2 } } \right) ^{ 4 }{ \left( \cfrac { -2 }{ { x }^{ 2 } } \right) }^{ 8 } }\right]$

$\quad \quad = \cfrac { 12! }{ 4!8! } .\cfrac { { x }^{ 8 } }{ { 2 }^{ 4 } } .\cfrac { { 2 }^{ 8 } }{ { x }^{ 16 } }$

$\quad \quad = { 7920x }^{ -4 }$

Consider the expansion of

$(1+x)^{2n+1}$
The average of the coefficients of the two middle terms in the expansion is

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

Explanation

Since,

$2n+1$ is odd.

Hence,

$\cfrac{2n+1+1}{2}$ and

$\cfrac{2n+1+3}{2}$ are two middle terms.

i.e.

$(n+1)th$ and

$(n+2)th$ terms are two middle terms.

$\therefore\cfrac{C^{2n+1}_n+ C^{2n+1}_{n+1}}{2}=\cfrac{C^{2n+1+1}_{n+1}}{2}$

$=\cfrac12C^{2n+2}_{n+1}=\cfrac12.\cfrac{2n+2}{n+1}C^{2n+1}_{n}=C^{2n+1}_{n}$

Hence, B is the correct option.

In the expansion of

$\left( x^3 - \dfrac{1}{x^2} \right )^n , n \in N$ , if the sum of the coefficient of

$x^5$ and

$x^{10}$ is

$0$ , then

$n$ is :

0%
$25$
0%
$20$
0%
$15$
0%
None of these

Explanation

Term of

${ x }^{ 5 }=^{ n }{ C }_{ r }\quad { x }^{ 3r }{ \left( \cfrac { -1 }{ { x }^{ 2 } } \right) }^{ n-r }=^{ n }{ C }_{ r }\quad { x }^{ 3r }. { x }^{ 2r-2n }. { \left( -1 \right) }^{ n-r }$
So

$3r+2r-2n=5$

$r=\cfrac { 5+2n }{ 5 } \longrightarrow \left( 1 \right)$
Also for

${ x }^{ 10 }$ be

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

${ x }^{ 10 }=^{ n }C_{ { r }^{ 1 } }{ x }^{ { 3r }^{ 1 } }{ \left( \cfrac { -1 }{ { x }^{ 2 } } \right) }^{ n-r }$

$\Rightarrow 3{ r }^{ 1 }+2{ r }^{ 1 }-2n=10$

${ r }^{ 1 }=\cfrac { 2n+10 }{ 5 } \longrightarrow \left( 2 \right)$
Now we know if

$\left| ^{ n }{ C }_{ r } \right| =\left| ^{ n }{ C }_{ { r }^{ 1 } } \right| \Rightarrow n=r+{ r }^{ 1 }$
Now we add co oefficient of

${ x }^{ 5 }\& { x }^{ 10 }$

$^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }+^{ n }{ C }_{ { r }^{ 1 } }{ \left( -1 \right) }^{ { n-r }^{ 1 } }=0$
Co oefficient of

${ x }^{ 5 }\quad is\quad ^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }$ ; co oeffiecient of

${ x }^{ 10 }\quad is\quad ^{ n }{ C }_{ { r }^{ 1 } }{ \left( -1 \right) }^{ { n-r }^{ 1 } }$

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

$\Rightarrow \left| ^{ n }{ C }_{ r } \right| =\left| ^{ n }{ C }_{ { r }^{ 1 } } \right|$
So,

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

$n=\cfrac { 2n+10 }{ 5 } +\cfrac { 2n+5 }{ 5 }$

$5n=4n+15$

$n=15$

The value of

${ }^nC_0 - { }^nC_1 + { }^n C_2 - .... + (-1)^{n^n}C_n$ is:

0%
$1$
0%
$0$
0%
$2^n$
0%
$n$

Explanation

By Binomial Expansion we know,

${ \left( a+b \right) }^{ n }=^{ n }{ C }_{ 0 }{ a }^{ 0 }{ b }^{ n }+^{ n }{ C }_{ 1 }{ a }^{ 1 }{ b }^{ n-1 }+....+^{ n }{ C }_{ n-1 }{ a }^{ n-1 }{ b }^{ 1 }+^{ n }{ C }_{ n }{ a }^{ n }{ b }^{ 0 }$
Now putting

$a=-1$ &

$b=+1$
we have,

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

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

The value of the expression

${ _{ }^{ k-1 }{ C } }_{ k-1 }+{ _{ }^{ k }{ C } }_{ k-1 }+....{ _{ }^{ n+k-2 }{ C } }_{ k-1 }$ is given by :

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

Explanation

We know,

$\displaystyle \sum _{ m=k }^{ n }{ { { ^{ m }{ C }_{ r}=\ ^{ n+1 }{ C }_{ r+1\\ } } } }$

According to the question,

$m= n+k-2$

$r= k-1$

Sum of series is given by

$= \ ^{ m+1 }{ C }_{ k+1\\ }$

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

How many terms are there in the expansion of

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

0%
$11$
0%
$20$
0%
$21$
0%
$30$

Explanation

Now,

$(1+2x+x^{ 2 })^{ 10 }=((1+x)^{ 2 })^{ 10 }=(1+x)^{ 20 }$

Now, the number of terms in the expansion of

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

$n+1$ .

Thus, the number of terms in the expansion of

$(1+x)^{20}$ will be

$20+1=21$ .

Hence, option C is correct.

Consider the expansion of

$(1+x)^{2n+1}$
The sum of the coefficients of all the terms in the expansion is

0%
$2^{2n-1}$
0%
$4^{n-1}$
0%
$2\times 4^n$
0%
None of the above

Explanation

We can get the sum of all coefficients by putting

$x=1$ in the expansion, because for calculating the coefficients we need the terms independent of

$x$ .

To find the sum of coefficients of all terms, put

$x=1$ in the given expression

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

$2^{2n+1}=2.2^{2n}=2.4^n$

Hence, C is the correct option.

What is n equal to ?

0%
$5$
0%
$10$
0%
$15$
0%
None of the above

Explanation

Using binomial theorem, we can write

$\displaystyle \left({ x }^{ 3 }-\cfrac { 1 }{x^{2}} \right)^{ n }=\sum_{ r=0 }^{ n }\ { _{ r }^{ n }{ C{ \left(x ^{ 3 }\right) }^{ n-r } } } { \left(\cfrac { -1 }{ { x }^{ 2 } } \right) }^{ r }\\ =\displaystyle \sum_{ r=0 }^{ n }{ { \left(-1\right) }^{ r }\ {}_{ r }^{ n }{ C }{ x }^{ \left(3n-5r\right) } }$

For

${x}^{3}$ and

${x}^{10}$ let value of r be

${ r }_{ 5 }$ and

${ r }_{ 10 }$ respectively

$3n-5{r}_{10}=5$ .........

$(i)$

$3n-5{r}_{10}=10$ ..........

$(ii)$

${ \left(-1\right) }^{ { r }_{ 5 } }\quad _{ { r }_{ 5 } }^{ n }{ C\quad }{ x }^{ \left(3n-5{ r }_{ 5 }\right) }$

$=-{ \left(-1\right) }^{ { r }_{ 10} }\quad _{ { r }_{ 10 } }^{ n }{ C\quad }{ x }^{ \left(3n-5{ r }_{ 10}\right) }$

Given sum of coefficient of

${x}^{5}$ and

${x}^{10}$ is 0

All terms are positive, but still there is a negative sign, means one of

${r}_{5}$ and

${r}_{10}$ is even and other is odd

By properties of combinations

$_{ { r }_{ 5 } }^{ n }{ C}$

$={}_{ { r }_{ 10 } }^{ n }{ C\quad }$

${r}_{5}+{r}_{10}=n$ ..........

$(iii)$

Adding

$(i)$ and

$(ii)$

$6n-5\left({r}_{5}+{r}_{10}\right)=15$

From

$(iii),$

${r}_{5}+{r}_{10}=n$

$\Rightarrow n=15$

In the expansion of

$\left (x + \dfrac {1}{x}\right )^{n}$ , then the coefficient of the term indepenent of x is

0%
$\dfrac {n!}{(r!)^{2}}$
0%
$\dfrac {n!}{(r + 1)! (r - 1)!}$
0%
$\dfrac {n!}{\left (\dfrac {n + r}{2}\right )! \left (\dfrac {n - r}{2}\right )!}$
0%
$\dfrac {n!}{\left [\left (\dfrac {n}{2}\right )!\right ]^{2}}$

Explanation

In the expansion of

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

$r^{th}$ term is

$t_{r+1}={}^{n}C_{r} x^{n-r} \left(\dfrac{1}{x}\right)^{r}$

$\implies t_{r+1}={}^{n}C_{r} x^{n-r} x^{-r}$

$\implies t_{r+1}={}^{n}C_{r} x^{n-2r}$ ...

$(i)$

We need to find the coefficient of

$x^{0}$

Then substitute

$n-2r=0\implies r=\dfrac{n}{2}$

Substitute

$r$ in RHS of

$(i)$ , we get

$t_{r+1}={}^{n}C_{\frac{n}{2}}x$

Then, coefficient of

$x$ is

$\displaystyle ={}^{n}C_{\frac{n}{2}}=\dfrac{n!}{\left(\dfrac{n}{2}\right)! \left(n-\dfrac{n}{2}\right)!}=\dfrac{n!}{\left(\dfrac{n}{2}\right)!\left(\dfrac{n}{2}\right)!}=\dfrac{n!}{\left[\left(\dfrac{n}{2}\right)!\right]^{2}}$

Hence, coefficient of

$x^{0}=x$ is

$\dfrac{n!}{\left[\left(\dfrac{n}{2}\right)!\right]^{2}}$

The sum of the series

$^{20}C_0 - \,^{20}C_1+\,^{20}C_2-\,^{20}C_3+...+\,^{20}C_{10}$ is

0%
$-^{20}C_{10}$
0%
$\dfrac{1}{2}\,^{20}C_{10}$
0%
$0$
0%
$^{20}C_{10}$

Explanation

On putting

$x=-1$ in

$(1+x)^{20}=^{20}C_0 + ^{20}C_1x+... +{20}C_{10}x^{10} + ... + \,^{20} C_{20} x^{20}$

We get,

$0=\,^{20}C_0 - \,^{20}C_1+... - \,^{20}C_9+\,^{20}C_{10} - \, ^{20}C_{11}+ ... + \,^{20}C_{20}$

$\Rightarrow 0 = \,^{20}C_0 - \,^{20}C_1+ ... - \,^{20}C_9 + \,^{20}C_{10}-\,^{20}C_9+...+\,^{20}C_0$ ........

$[\because {}^{n}C_{r}={}^{n}C_{n-r}]$

$\Rightarrow 0 = 2(^{20}C_0-\,^{20}C_1+ ... -\,^{20}C_9)+\,^{20}C_{10}$

$\Rightarrow \,^{20}C_{10} = 2(^{20}C_0 - \,^{20}C_1+ ... + \,^{20}C_{10})$ .... [Adding

${}^{20}C_{10}$ on both the sides]

$\Rightarrow \,^{20}C_0 - \,^{20}C_1+...+\,^{20}C_{10}=\dfrac{1}{2}\,^{20}C_{10}$

$T_r=^{2016}C_rx^{2016-r}$ , for

$r=0, 1, ,....2016$ , then

$(T_0 - T_2+T_4....+T_{2016})^2+(T_1-T_3+T_5....T_{2015})^2$ is equal to -

0%
$(X^2-1)^{1008}$
0%
$(X+1)^{2016}$
0%
$(X^2-1)^{2016}$
0%
$(X^2+1)^{2016}$

Explanation

Let

$i$ represent iota.

$(T_0-T_2+T_4+\dots+T_{2016}) = (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})\dots(1)$

Now, multiply

$(T_1-T_3+T_5+\dots-T_{2015})$ with

$-i$ to get

$-(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})\dots(2)$

$\implies (T_0-T_2+T_4+\dots+T_{2016})^2+(T_1-T_3+T_5+\dots-T_{2015})^2$

$= (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 + i^4(T_1-T_3+T_5+\dots-T_{2015})^2$

$= (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 -(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})^2$

$= (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$ (As

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

Now, consider these terms seperately

$(T_0+iT_1+i^2T_2+\dots)=(^{2016}C_0x^{2016}i^0 +^{2016}C_1x^{2015}i^1\dots) = (i+x)^{2016}$

$(T_0-iT_1+i^2T_2-i^3T_3\dots) = (^{2016}C_0(ix)^{2016} +^{2016}C_1(ix)^{2015}\dots) = (1+ix)^{2016}$

$\therefore (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$

$= (i+x)^{2016}\times(1+ix)^{2016}$

$= [(i+x)(1+ix)]^{2016}$

$= [i-x+x+ix^2]^{2016}$

$= (1+x^2)^{2016}$

$C_{0}, C_{1}, C_{2}, ...., C_{n}$ are binomial coefficients of order

$n$ , then the value of

$\dfrac {C_{1}}{2} + \dfrac {C_{3}}{4} + \dfrac {C_{5}}{6} + .... =$

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

Explanation

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

Integrating, We have

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

Putting,

$x=1$

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

Putting,

$x=-1$

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

Adding above two equations we have,

$\dfrac { (2)^{ n+1 } }{ n+1 } =2\left(\dfrac { C_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } .....\right)$

$\implies \dfrac { (2)^{ n } }{ n+1 } =\dfrac { C_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } .....$

Hence, option D is correct.

The total number of terms in the expansion of

$(x + a)^{47} - (x - a)^{47}$ after simplification is

0%
24
0%
47
0%
48
0%
96

Explanation

${ \left( x+a \right) }^{ 47 }-{ \left( x-a \right) }^{ 47 }$

When we expand the above equation using binomial expansion

$(x +y)^{n} = \displaystyle (\sum_{k=0}^{n} {^{n}C_k} x^{k}y^{n-k})$

So the above equation becomes

$(x +a)^{47} = \displaystyle (\sum_{k=0}^{47} {^{47}C_k} x^{k}a^{47-k})$

$(x -a)^{47} = \displaystyle (\sum_{k=0}^{47} {^{47}C_k} x^{k}(-a)^{47-k})$

${ \left( x+a \right) }^{ 47 }\Rightarrow$ There are 48 terms in the expansion and all are positive

${ \left( x-a \right) }^{ 47 }\Rightarrow$ There are 48 terms in the expansion

The terms with odd powers of a will be cancelled and those with even powers of a will add up.
24 terms will be positive and 24 negative in the expansion of

$(x-a)^{47}$
48 terms positive-[24 terms negative and 24 terms positive]

$=48\quad terms\quad positive +24\quad terms\quad negative+24\quad terms\quad positive$

$=24$ terms

Let

$((1 + x) + x^{2})^{9} = a_{0} + a_{1}x + a_{2}x^{2} + ..... + a_{18}x^{18}$ . Then

0%
$a_{0} + a_{2} + ..... + a_{18} = a_{1} + a_{3} + ..... + a_{17}$
0%
$a_{0} + a_{2} + ..... + a_{18}$ is even
0%
$a_{0} + a_{2} + ..... + a_{18}$ is divisible by $9$
0%
$a_{0} + a_{2} + ..... + a_{18}$ is divisible by $3$ but not by $9$

Explanation

Given :

$((1 + x) + x^{2})^{9} = a_{0} + a_{1}x + a_{2}x^{2} + ..... + a_{18}x^{18}$ .....

$(i)$

Put

$x=1$ in

$(i)$ , we get

$(1+1+1)^{9}=a_{0}+a_{1}+a_{2}+.....+a_{18}$

${3}^{9}=a_{0}+a_{1}+a_{2}+.....+a_{18}$ ....

$(ii)$

Put

$x=-1$ in

$(i)$ , we get

$(1-1+1)^{9}=a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-..... -a_{17}+a_{18}$

$1=a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-..... -a_{17}+a_{18}$ .....

$(iii)$

Adding

$(ii)$ and

$(iii)$ , we get

$3^{9}+1=a_{0}+a_{1}+....+a_{18}+a_{0}-a_{1}+a_{2}-.....-a_{17}+a_{18}$

$=2a_{0}+2a_{2}+2a_{4}+....+2a_{18}$

$\implies 2(a_{0}+a_{2}+a_{4}+....+a_{18})=3^{9} + 1$

$\implies a_{0} + a_{2} + ..... + a_{18} = \dfrac {3^{9} + 1}{2} \rightarrow even$

Hence,

$a_{0}+a_{2}+a_{4}+....+a_{18}$ is even.

Let

$n \ge 5$ and

$b \neq 0$ . In the binomial expansion of

${ \left( a-b \right) }^{ n }$ , the sum of the 5th and 6th terms is zero then

${ a }/{ b }$ equals

0%
$\dfrac { 5 }{ n-4 }$
0%
$\dfrac { 1 }{ 5\left( n-4 \right) }$
0%
$\dfrac{n-5}{6}$
0%
$\dfrac{n-4}{5}$

Explanation

Solution:

Given that:

$n\ge5,b\neq0,$ given expansion is

$(a-b)^n$ and

$t_5+t_6=0$

To find:

$\cfrac ab=?$

Solution:

$t_5={}^nC_4a^4(-b)^{(n-4)}$ and

$t_6={}^nC_5a^5(-b)^{(n-5)}$

$\because t_5+t_6=0$

$\therefore {}^nC_4a^4(-b)^{(n-4)}$

$+{}^nC_5a^5(-b)^{(n-5)}=0$

$\Longrightarrow {}^nC_5a={}^nC_4b$

$\Longrightarrow \cfrac ab=\cfrac{{}^nC_4}{{}^5C_5}=\cfrac{\cfrac{n!}{4!(n-4)!}}{\cfrac{n!}{5!(n-5)!}}=\cfrac{5}{n-4}$

Hence, A is the correct answer.

In the expansion of

${ \left( 3x-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 10 }$ , the

$5^{th}$ term from the end is

0%
$\cfrac { 16486 }{ { x }^{ 8 } }$
0%
$\cfrac { 17010 }{ { x }^{ 8 } }$
0%
$\cfrac { 13486 }{ { x }^{ 8 } }$
0%
None of these

Explanation

There are

$11$ terms in the expansion of

${ \left( 3x-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 10 }$
Therefore,

$5^{th}$ term from the end.

$=(10-5+2)^{th}$ term from beginning

$={ T }_{ 7 }={ _{ }^{ 10 }{ C } }_{ 6 }{ \left( 3x \right) }^{ 10-6 }{ \left( -\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 6 }$

$=\cfrac { 10! }{ 6!4! } { 3 }^{ 4 }{ x }^{ 4 }\cfrac { 1 }{ { x }^{ 12 } }$

$=\cfrac { 17010 }{ { x }^{ 8 } }$

The coefficient of

$x^{49}$ in the product

$(x - 1) (x - 2) (x - 3) .... (x - 50)$ is

0%
$-2250$
0%
$-1275$
0%
$1275$
0%
$2250$
0%
$-49$

Explanation

Coefficient of

$x^{49}$ in product of

$(x-1)(x-2)(x-3)(x-4)........(x-50)$

We know that,

$(x-1)(x-2)(x-3)(x-4)........(x-n)=x^n-(1+2+3+....+n)x^{n-1}+.......$

Coefficient of

$x^{49}=-(1+2+3+...+50)$

$=-\cfrac{50\times51}{2}$

$=-1275$

Hence, B is the correct option.

$C_0, C_1, C_2, C_3, ....$ are binomial coefficients in the expansion of

$( 1+x)^n$ , then

$\dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac{}{} - \dfrac{}{} + ...$ is equal to :

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

Explanation

We know

$(1-x)^n = C_0 - C_1 x + C_2x^2 - ... + (-1)^n C_n \cdot x^n$

On multiplying both sides bby

$x^2$ , we get

$(1-x)^n x^2 = C_0x^2 - C_1 x^3 + C_2x^4 - ... +$

On integrating both sides by taking limit

$0$ to

$1$ .

Therefore,

$\displaystyle \int_0^1 ( 1-x)^n x^2 dx =\displaystyle \int_0^1 ( C_0x^2 - C_1x^3 + C_2x^4 + ....) dx$

$\displaystyle \int_0^1 x^n (1-x)^2 dx = \left[ C_0 \dfrac{x^3}{3} - C_1 \dfrac {x^4}{4} + C_2 \dfrac{x^5}{5} - ... \right]_0^1$

$\Rightarrow\displaystyle \int_0^1 x^n ( 1 + x^2 - 2x) dx = \dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac {C_2}{5} - ...$

Here

$\dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac {C_2}{5} - ...$

$= \left[ \dfrac{x^{n+1}}{n+1} + \dfrac{x^{n+3}}{n+3} - \dfrac{2x^{n+2}}{n+2} \right]_0^1$

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

The value of

$r$ for which the coefficients of

$(r-5)$ th and

$(3r+1)$ th terms in the expansion of

${(1+x)}^{1/2}$ are equal, is

0%
$4$
0%
$9$
0%
$12$
0%
None of these

Explanation

Since, coefficient of

$(r-5$ th term

$=$ coefficient of

$=(3r+1)$ th term

$\quad { _{ }^{ 12 }{ C } }_{ r-6 }=\quad { _{ }^{ 12 }{ C } }_{ 3r }$

$\Rightarrow$

$r-6=3r$
or

$12-r+6=3r$

$\rightarrow$

$2r=-6$ or

$4r=18$

$\Rightarrow$

$r=-3$ or

$r=\cfrac { 18 }{ 4 }$
Hence no value of

$r$ exist, because

$r$ neither be negative nor in fraction

$(1 + x + x^{2})^{n} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{2n}x^{2n}$ , then

$2a_{1} - 3a_{2} + ... -(2n + 1)a_{2n}$ is equal to

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

Explanation

Given,

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

$\Rightarrow x(1 + x + x^{2})^{n} = x + a_{1}x^{2} + a_{2}x^{3} + ... + a_{2n}x^{2n + 1}$

On diffferentiating w.r.t.

$x$ , we get

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

$= 1 + 2a_{1}x + 3a_{2}x^{2} + ... + a_{2n} \cdot (2n + 1)x^{2n}$

On putting

$x = -1$ , we get

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

$= 1 - 2a_{1} + 3a_{2} + ... + a_{2n} (2n + 1)$

$\Rightarrow 1 - n(-1) = 1 - 2a_{1} + 3a_{2} + ... + a_{2n}(2n + 1)$

$\Rightarrow 2a_{1} - 3a_{2} .... =(2n + 1)a_{2n} = -n$

The middle term in the expansion of

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

0%
$^{10}C_5$
0%
$^{10}C_6$
0%
$^{10}C_5 \dfrac{1}{x^{10}}$
0%
$^{10}C_5 x^{10}$
0%
$^{10}C_5 10^{10}$

Explanation

MIddle term of the expansion of

$\left(\cfrac{10}x+\cfrac x{10}\right)^{10}$ is

$T_6.$

$T_6={}^{10}C_{5}\left(\cfrac{10}x\right)^5\left(\cfrac x{10}\right)^5$

$T_6={}^{10}C_{5}$

Hence, A is the correct option.

If the term free from

$x$ in the expansion of

$\left (\sqrt {x} - \dfrac {k}{x^{2}}\right )^{10}$ is

$405$ , then the value of

$k$ is

0%
$\pm 1$
0%
$\pm 3$
0%
$\pm 4$
0%
$\pm 2$

Explanation

General term in the expansion of

$\left (\sqrt {x} - \dfrac {k}{x^{2}}\right )$

$T_{r + 1} = ^{10}C_{r} (\sqrt {x})^{10 - r} \left (\tfrac {-k}{x^{2}}\right )^{r}$

$= ^{10}C_{r}x^{\dfrac {10 - r}{2}} \cdot (-k)^{r} \cdot x^{-2r}$

$= ^{10}C_{r} (-k)^{r} x^{\left (\dfrac {10 - 5r}{2}\right )}$

The term is free from

$x$ .

Put

$\dfrac {10 - 5r}{2} = 0$

$\Rightarrow r = 2$

Now,

$^{10}C_{2} (-k)^{2} = 405$

$\Rightarrow \dfrac {10\times 9}{1\times 2}\cdot k^{2} = 405$

$\Rightarrow k^{2} = \dfrac {405}{45} = 9$

$\Rightarrow k = \pm 3$ .

Sum of coefficients of the last

$6$ terms in the expansion of

${ \left( 1+x \right) }^{ 11 }$ when the expansion is in ascending powers of

$x$ , is

0%
$2048$
0%
$32$
0%
$512$
0%
$64$
0%
$1024$

Explanation

$(1+x)^{\prime \prime}=\sum_{r=0}^{11} n_{C_{r}}(1)^{n-r}(x)^{r} \\$

$\text { last } 6 \text { terms of this expansion are } \\$

$\qquad{ }^{\prime \prime} C_{11}+{ }^{\prime \prime} C_{10}+{ }^{\prime \prime} C_{q}+{ }^{\prime \prime} C_{8}+{ }^{\prime \prime} C_{7}+{ }^{\prime \prime} C_{6} \\$

$\text { using }{ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}$

$=12 C_{11}+{ }^{12} C_{9}+{ }^{12} C_{7} \\$

$=12+220+792 \\$

$=1024$

${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },\dots ,{ C }_{ 15 }$ are binomial coefficients in

${ \left( 1+x \right) }^{ 15 }$ , then

$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\dfrac { { C }_{ 3 } }{ { C }_{ 2 } } +\cdots +15\dfrac { { C }_{ 15 } }{ { C }_{ 14 } }$ is equal to

0%
$60$
0%
$120$
0%
$64$
0%
$124$
0%
$144$

Explanation

We know that,

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

$\Rightarrow r\cdot \dfrac { ^{ n }{ { C }_{ r } } }{ ^{ n }{ { C }_{ r-1 } } } =n+1-r$

$\Rightarrow \displaystyle\sum _{ r=1 }^{ n }{ r\cdot \dfrac { ^{ 16 }{ { C }_{ r } } }{ ^{ 16 }{ { C }_{ r-1 } } } } =\displaystyle\sum _{ r=1 }^{ 16 }{ \left( 16-r \right) }$

$\displaystyle 16\times 15-\sum_{r=1}^{n}r$

$=16\times 15-\dfrac { 15\times 16 }{ 2 }$

$=240-120$

$=120$

The middle term in the expansion of

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

0%
$\cfrac { 1.3.5....(2n-1){ 2 }^{ n } }{ n! }$
0%
$\cfrac { 1.2.3....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! }$
0%
$\cfrac { 1.3.5....(2n-1){ x }^{ n } }{ n! }$
0%
$\cfrac { 1.3.5....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! }$

Explanation

Given expansion is

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

The total number of terms are

$2n+1$ which is an odd number,

$\therefore$ there is only one middle term which is

$\left( \begin{matrix} 2n \\ n \end{matrix} \right)$ ,

$\left( \begin{matrix} 2n \\ n \end{matrix} \right) =\dfrac { \left( 2n \right) ! }{ n!n! } \\$

$\Longrightarrow \dfrac { \left[ 1.3.5...\left( 2n-1 \right) \right] \left[ { 2 }^{ n }n! \right] }{ n!n! } { x }^{ n }=\dfrac { \left[ 1.3.5...\left( 2n-1 \right) \right] \left[ { 2 }^{ n } \right] }{ n! } { x }^{ n }$

$n\epsilon N$ and

$(1 + 4x + 4x^{2})^{n} = \displaystyle \sum_{r = 0}^{r = 2n} a_{r}x^{r}$ then value of

$2\displaystyle \sum_{r = 0}^{n}a_{2r}$ equals

0%
$9^{n} - 1$
0%
$9^{n} + 1$
0%
$3^{n }+ 1$
0%
$3^{n} - 1$

Explanation

$\because 1 + 4x + 4x^{2} = (1 + 2x)^{2}$

$\therefore (1 + 2x)^{2n} = \displaystyle \sum_{r = 0}^{2n}a_{r}x^{r}$
putting

$x = 1$

$3^{2n} = \displaystyle \sum_{r = 0}^{2n} a_{r} = a_{0} + a_{1} + a_{2} + a_{3} + ..... + a_{2n} .... (i)$
and putting

$x = -1$

$1 = \displaystyle \sum_{r = 0}^{r = 2n}a_{r} (-1)^{r} = a_{0} - a_{1} + a_{2} - a_{3} + .... + a_{2n} ... (ii)$
Now adding (i) and (ii) we get

$2(a_{0} + a_{2} + a_{4} + .... + a_{2n}) = 3^{2n} + 1$

$\Rightarrow 2\displaystyle \sum_{r = 0}^{r =n} a_{2r} = 9^{n} + 1$
Hence choice (b) is correct answer.

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

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

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

Explanation

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

$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 -------(i)$

$(a-b)^5=a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5 -------(ii)$

$(a+b)^5+(a-b)^5=2[a^5+10a^3b^2+5ab^4]$

$=2[x^5+10x^3(x^3-1)+5x(x^3-1)^4]$

$=2[x^5+10x^6-10x^3+5x(x^6-2x^3+1)]$

$=2x^5+20x^6-20x^3+10x^7-20x^4+10x$

Here all the co-efficient of the above equation are odd terms coefficients of even terms are cancelled out.

So, sum of coefficients

$\Rightarrow2+20-20+10-20+10=2$

${C}_{r}$ denotes the binomial coefficient

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

$\left( -1 \right) { C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+......(3n-1){ C }_{ n }^{ 2 }=\quad$

0%
$\left( 3n-2 \right) { _{ }^{ 2n }{ C } }_{ n }$
0%
$\left( \cfrac { 3n-2 }{ 2 } \right) { _{ }^{ 2n }{ C } }_{ n }$
0%
$\left( 5+3n \right) { _{ }^{ 2n }{ C } }_{ n }$
0%
$\left( \cfrac { 3n-5 }{ 2 } \right) { _{ }^{ 2n }{ C } }_{ n+1 }$

Explanation

The general term of above series is

$T_{r} = (3r-1)C_{r}^{2}$

$(-1)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+(3n-1)C_{n}^{2}$

$=\sum_{r=0}^{r=n}(3r-1)C_{r}^{2}$

$=3\sum_{r=0}^{r=n}rC_{r}^{2} - \sum_{r=0}^{r=n}C_{r}^{2}$

$=3S_{1} - S_{2}$

The binomial expansion of

$(1+x)^n$ and

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

$(1+x)^n = C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}$ ....[1]

$(1+\dfrac{1}{x})^n = C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+....+C_{n}\dfrac{1}{x^{n}} = \sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}$ ....[2]

To find the value of

$S_{2}$ , we can multiply above two equations and then compare the coefficients of terms independent of x.

Multiplying equation [1] and [2], we get

$(1+x)^n (1+\dfrac{1}{x})^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}$

$\implies \dfrac{(1+x)^{2n}}{x^{n}} = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}$

$\implies \dfrac{(1+x)^{2n}}{x^{n}}=C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}$ + terms containing x

Therefore, comparing coefficients of

$x^{0}$ in L.H.S and R.H.S, we get

Coefficients of

$x^{0}$ in R.H.S = Coefficients of

$x^{0}$ in L.H.S

$\implies C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}=$ coefficient of

$x^{0}$ in

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

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

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

Differentiating with respect to x, we get

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

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

To find the value of

$S_{1}$ , we can multiply equations [1] and [4], and then compare the coefficients of terms independent of x.

Multiplying equation [1] and [4], we get

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

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

Therefore, comparing coefficients of

$x^{0}$ in L.H.S and R.H.S, we get

Coefficients of

$x^{0}$ in R.H.S = Coefficients of

$x^{0}$ in L.H.S

$\implies C_{1}^{2}+2C_{2}^{2}+3C_{3}^{2}+...+nC_{n}^{2}=$ coefficient of

$x^{0}$ in

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

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

Therefore,

$3S_{1} - S_{2} = 3n(^{2n-1}C_{n})-^{2n}C_{n}$

$= 3n\dfrac{(^{2n}C_{n})}{2}-^{2n}C_{n}$

$= (\dfrac{3n-2}{2})^{2n}C_{n}$

Hence, the answer is option (B)

Given

$(1-2x+5x^2-10x^3)(1+x)^n=1+a_1x+a_2x^2+...$ and that

$a_1^2=2a_2$ then the value of

$n$ is-

0%
6
0%
2
0%
5
0%
3

Explanation

Given that

$\left( 1-2x+5{ x }^{ 2 }-10{ x }^{ 3 } \right) { \left( 1+x \right) }^{ n }=1+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+...$

Given that

${ { a }_{ 1 } }^{ 2 }=2{ a }_{ 2 }$

Here

${ a }_{ 1 }$ is the coefficient of

$x$ and

${ a }_{ 2 }$ is the coefficient of

${ x }^{ 2 }$ ,

$\Longrightarrow { a }_{ 1 }=n-2$ and

$\Longrightarrow { a }_{ 2 }=-2n+{ C }_{ 2 }+5$

Substituting these values in the given relation we get,

$\Longrightarrow { \left( n-2 \right) }^{ 2 }=2\left( 5-2n+\frac { n\left( n-1 \right) }{ 2 } \right) \\ \Longrightarrow n=6\\$

${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },.....{ C }_{ r }$ are binomial coefficients in the expansion of

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

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

0%
$\sum _{ r=1 }^{ n }{ { \left( -1 \right) }^{ n-1 } } \cfrac { { C }_{ r } }{ r }$
0%
$\sum _{ r=1 }^{ n }{ \cfrac { 1 }{ r } }$
0%
$\sum _{ r=2 }^{ n }{ \cfrac { 1 }{ r-1 } }$
0%
None of these

Explanation

Consider

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

$\Rightarrow 1-{ (1-x) }^{ n }={ C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-......$

integrating both side w.r.t

$x$ with limit

$0$ to

$1$

$\displaystyle \therefore \int _{ 0 }^{ 1 }{ \left( { C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-..... \right) dx } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ (1-x) }^{ n } }{ 1-(1-x) } } dx$

$\displaystyle \Rightarrow \cfrac { { C }_{ 1 } }{ 1 } -\cfrac { { C }_{ 2 } }{ 2 } +\cfrac { { C }_{ 3 } }{ 3 } +....{ (-1) }^{ n }-1\cfrac { { C }_{ n } }{ n } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ x }^{ n } }{ 1-x } } dx\quad$

$\displaystyle \left( Using\quad \int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a- } x)dx \right)$

$\displaystyle =\int _{ 0 }^{ 1 }{ \left( 1+x+{ x }^{ 2 }+...+{ x }^{ n-1 } \right) dx }$

$=1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +...\cfrac { 1 }{ n } =\sum _{ r=1 }^{ n }{ \cfrac { 1 }{ r } }$

Prove that the coefficient of middle term in the expansion of

$( 1 + x ) ^{2n}$ is equal to the sum of the coefficient of two middle terms in

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

0%
True
0%
False

Explanation

Statement is true, so option

$A$ is correct.

Proof :- Middle term of

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

Since

$2n$ is even middle term =

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

$T_{n + 1}$

we know that general & term of expansion

$(a + b)^n$ is -

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

here,

$r = n , n = 2n , a = 1 , b = x$

putting values

$T_{nH} = \, ^{2n}C_n (1)^{2n - n} (x)^n$

$T_{nH} = \,^{2n}C_n (x)^n......(A)$

Here, coefficient is

$^{2n} C_n$

$\Rightarrow$ middle term of

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

Since,

$(2n - 1)$ is add

There will be 2 middle terms

$\dfrac{(2n - 1) + 1}{2}$ and

$\left(\dfrac{(2n - 1) + 1}{2} + 1 \right)$ term

$= n^{th}$ term and

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

$= T_n$ and

$T_{n + 1}$

we know that general term for expansion

$(a + b)^n$

$T_{rH} = \,^nC_r \, a^{n - r} \, b^r$

for

$T_n$ in

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

Putting

$r = n - 1, n = 2n - 1, a = 1 \, \& \, b = x$

$T_n = \, ^{2n - 1} C_{n - 1} \, (1)^{2n - 1 - ( n - 1)} x_{( n- 1)} = \, ^{2n - 1}C_{n - 1} x^{(n - 1)}$

Coefficient pf middle term

$(n^{th} \, term) = \, ^{2n - 1}C_{n - 1}$

$T_{nH} = \, ^{2n - 1}C_n \, (1)^{2n - 1 - n} x^{(n)} = \, ^{2n - 1} C_{n} \, (X)^n$

Sum of the coefficent of the middle term in the expansion of

$(1 + X)^{2n - 1}$

$= \,^{2n -1}C_{n - 1} + \, ^{2n - 1}C_n$

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

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

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

$= \dfrac{2n (2n - 1)!}{n! n (n - 1)!} = \dfrac{(2n)!}{n! \, n!} = \, ^{2n}C_n .....(B)$

Equation

$(A)$ = Equation

$(B)$

State true or false.

The middle term in the expansion of

$\left(\dfrac{x}{a} \, - \, \dfrac{a}{x}\right)^{10}=-252$

0%
True
0%
False

Find the term of the expansion of

$\displaystyle\, \left ( \sqrt[3]{x^{-2}} + x \right )^7$ containing

$x$ in the second power.

0%
$T_4$
0%
$T_5$
0%
$T_6$
0%
$T_7$

Explanation

Note:- The general term or

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

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

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

${T}_{r+1}=^{ 7 }{ C }_{ r }\left( \sqrt [ 3 ]{ { x }^{ -2 } } \right) ^{ 7-r }\left(x \right)^{r}$

$\Rightarrow {T}_{r+1}= ^{ 7 }{ C }_{ r }\left({x}^{-2}\right)^{{7-r}/{3}}{x}^{r}$

$\Rightarrow {T}_{r+1}=^{ 7 }{ C }_{ r } {x}^{{-2}/{3}(7-r)+r}$

Now, if power of

$x$ must be

$2$ , then

$r$ is:-

$\therefore {x}^{{-2}/{3}(7-r)+r}= {x}^{2}$

$\Rightarrow \cfrac{-2}{3}(7-r)+r=2$

$\Rightarrow \cfrac{-14}{3}+\cfrac{2}{3}r+r=2$

$\Rightarrow \cfrac{5r}{3}= \cfrac{20}{3}$

$\Rightarrow r=4$

Therefore

${T}_{4+1}={T}_{5}$ is the term which contains

${x}^{2}$

Sum of coefficients in the expression of

$(x+2y+z)^{10}$ is

0%
$2^{10}$
0%
$3^{10}$
0%
1
0%
None of these

Explanation

Sum of = coeff.

$4^{10}$

The

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

$\left (2x^2 - \dfrac {1}{3x^2}\right)^{10}$

0%
$-\dfrac{986}{27}$
0%
$\dfrac{986}{27}$
0%
$\dfrac{896}{27}$
0%
$- \dfrac{896}{27}$

Explanation

${ \left( 2{ x }^{ 2 }-\dfrac { 1 }{ { { 3x }^{ 2 } } } \right) }^{ 10 }$

$6^{th}$ coefficient

$\rightarrow$

$5^{th}$ term

$={ ^{ 10 }{ C } }_{ 5 }{ \left( 2{ x }^{ 2 } \right) }^{ 5 }{ \left( \dfrac { -1 }{ { { 3x }^{ 2 } } } \right) }^{ 5 }$

$={ ^{ 10 }{ C } }_{ 5 }\times 2^5\times \dfrac{-1}{3^5}$

$=-\dfrac{252\times 32}{243}$

$=-\dfrac{896}{27}.$

Hence, the answer is

$-\dfrac{896}{27}.$

Prove that

$C_0+C_1+C_2+.....C_n=2^n$

0%
$2^n$
0%
$2^{n-1}$
0%
$2^{n+1}$
0%
None of the above.

Explanation

$(1+x)^n=^nC_0+^nC_1x+\cdots+^nC_nx^n$

Let

$x=1$

$\Rightarrow C_0+C_1.1+C_2.1^2+\cdots+C_n.1^n=(1+1)^n=2^n$

$\Rightarrow C_0+C_1+C_2.1+\cdots+C_n=(1+1)^n=2^n$

In the expansion of

$(\sqrt[5]{3}+\sqrt[7]{2})^{24}$ , the rational term is

0%
$T_{14}$
0%
$T_{16}$
0%
$T_{15}$
0%
$T_7$

the coefficients of

$x^{49}$ in the polynomial.

$\left (x \, - \, \dfrac{C_1}{C_0}\right) \, \left (x \, - \, 2^2 \, \dfrac{C_2}{C_1}\right) \, \left (x \, - \, 3^2 \, \dfrac{C_3}{C_2}\right) \, ..... \, \, \left (x \, - \, 50^2 \, \dfrac{C_{50}}{C_{49}}\right)$ is

0%
$50\displaystyle \sum _{ r=1 }^{ 50 }{ r-{ r }^{ 2 } }$
0%
$51\displaystyle \sum _{ r=1 }^{ 50 }{ { r }^{ 2 } } -\displaystyle \sum _{ r=1 }^{ 50 }{ r }$
0%
$51\displaystyle \sum _{ r=1 }^{ 50 }{ { r } } -\displaystyle \sum _{ r=1 }^{ 50 }{ r^{ 2 } }$
0%
$none\ of\ these$

Explanation

We Know that

$(x \, - \, a) \, (x \, - \, b) \,(x \, - \, c).....50$
factors

$= \, x^{50}\, x^{49} \, . S_1 \, + x^{48} \, S_2 \, -....$
The coefficient of

$x^{49}$ is

$S_1 \, = \,- \,\left [ \dfrac{C_1}{C_0} \, + \, 2^2 \, \dfrac{C_2}{C_1} \,+ \,3^2 \, \dfrac{C_3}{C_2}\right] \, + .....r^2 \,\dfrac{C_r}{C_{r \, - \, 1}}]$

$Now \, r^2 \, .\dfrac{C_r}{C_{r \, - \, 1}} \, = \, r^2 \, .\dfrac{n \, -r \, + \, 1}{r}$

$= r \, (51 \, - \, r) \, = \, 51r \, - r^2$

$\, \therefore \, \sum_{r \, = \, 1}^{50} \, r^2 \, \dfrac{c_r}{C_{r \, - \, 1}} \, 51 \, \sum_{r \, = \, 1}^{50} \, r \, - \, \sum_{r \, = \, 1}^{50} \, r^2$

$= \, 51. \, \dfrac{N (N \, + \, 1)}{2}\, - \, \dfrac{N(N \, + \, 1)\, (2N \, + \, 1)}{6}$

$= \, 51. \, \dfrac{50. \, 51)}{2}\, - \, \dfrac{50. \ 51. \, (101)}{6}$

$= \,\dfrac{1}{6} \, 50. \, 51 \, \left [ 3 \, \times \, 51 \, - \, 101 \right ] \, = \, 25 \, \times \, 17 \, \times \, 52$

If the sum of the co-efficient in the expansion of

$(a+b)^n$ is

$1024$ , then the greatest co-efficient in the expansion is

0%
$252$
0%
$352$
0%
$452$
0%
$552$

Explanation

Sum of coefficient in the expansion of

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

$1024$

put

$a=b=1,2^{n}=1024\implies n=10$

The greatest coefficient is

$^{10}C_{5}=252$

$n\ge 2$ then

$3.{ C }_{ 1 }-4.{ C }_{ 2 }+5.{ C }_{ 3 }-......+{ \left( -1 \right) }^{ n-1 }\left( n+2 \right) .{ C }_{ n }$ is equal to

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

Explanation

$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+....+^nC_n x^n$

$x^2(1+x)^n=^nC_0 x^2+^nC_1x^3+^nC_2x^4+....+^nC_n x^{n+2}$

differentiate w.r.t

$x$

$2x(1+x)^n+nx^2(1+x)^{n-1}=2^nC_0 x+3 ^nC_1x^2+4 ^nC_2x^3+....+(n+2)^nC_n x^{n+1}$

put

$x=-1$

$0=-2+3 ^nC_1-4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n$

$3 ^nC_1- 4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n=2$

The sum

$^{10}C_3 + ^{11}C_3 + ^{12}C_3 + .... + ^{20}C_3$ is equal to

0%
$^{21}C_4$
0%
$^{21}C_4 - ^{10}C_4$
0%
$^{21}C_4 - ^{11}C_4$
0%
$^{21}C_{17}$

Explanation

$^{10}C_{3} +\ ^{11}C_{3} +\ ^{12}C_{3} + --- +\ ^{20}C_{3}$ ....(i)

addition and subtraction

$^{10}C_{4}$ in equation (i)

$^{10}C_{4} -\ ^{10}C_{4} +\ ^{10}C_{3} +\ ^{11}C_{3} + --- +\ ^{20}C_{3}$

[we know that -

$^nc_{r-1} +\ ^nc_{r} =\ ^{n+1}c_{r}]$

$^{10}C_{3} +\ ^{10}C_{4} +\ ^{11}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}$

$=\ ^{11}C_{4} +\ ^{11}C_{3} +\ ^{12}C_{3} + ---+\ ^{20}C_{3} -\ ^{10}C_{4}$

$=\ ^{12}C_{4} +\ ^{12}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}$

$=\ ^{13}C_{4} +\ ^{14}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}$

Similarly, Continue

So, we get

$^{21}C_{4} - \ ^{10}C_{4}$

So, option (B) is correct

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 7 - 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 7 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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