MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 6

The sum of all the coefficients in the binomial expansion of

$(x^2 + x^3)^{319}$ is

0%
1
0%
$(2)^{318}$
0%
$(2)^{319}$
0%
0

Explanation

For the sum of the coefficients of the binomial theorem, we substitute

$x=1$ .

Hence, we get

$(x^{2}+x^{3})^{319}|_{x=1}$

$=2^{319}$

Thus sum of the coefficients of the above binomial expansion is

$2^{319}$ .

$^{10}C_1+^{10}C_3+^{10}C_5+^{10}C_7+^{10}C_9=$

0%
$2^9$
0%
$2^{10}$
0%
$2^{10}-1$
0%
None of these

Explanation

For

$(X+A)^{ n }={ { C }_{ 0 }^{ n }X }^{ n }+\cdots{ C }_{ r }^{ n }{ X }^{ n-r }{ A }^{ r }+\cdots{ C }_{ n }^{ n }A^{ n }$

Put

$X=1; A=-1$

Hence,

$0= { C }_{ 0 }^{ n }-{ C }_{ 1 }^{ n }+{ C }_{ 2 }^{ n }-{ C }_{ 3 }^{ n }\cdots$ and so on

$\Rightarrow{ C }_{ 0 }^{ n }+{ C }_{ 2 }^{ n }{ +C }_{ 4 }^{ n }+\cdots={ C }_{ 1 }^{ n }+{ C }_{ 3 }^{ n }+{ C }_{ 5 }^{ n }+\cdots$ ...(1)

Now put

$A=1$ and we see that

${ 2 }^{ n }={ C }_{ 0 }^{ n }+{ C }_{ 1 }^{ n }+{ C }_{ 2 }^{ n }{ +{ C }_{ 3 }^{ n }+C }_{ 4 }^{ n }+{ C }_{ 5 }^{ n }+\cdots$ ...(2)

Hence, the sum of

${ C }_{ 1 }^{ n }+{ C }_{ 3 }^{ n }+{ C }_{ 5 }^{ n }+\cdots={ 2 }^{ n-1 }$ (From (1) and (2))

In the problem,

$n=10$

Hence, answer is

${ 2 }^{ 9 }$

The value of

$3{\;}^nC_0-8{\;}^nC_1+13 {\;}^nC_2-18 {\;}^nC_3+.....+n$ terms is

0%
$0$
0%
$3^n$
0%
$5^n$
0%
none of these

Explanation

The general term of the series is

$Tr = (1)^r (3 + 5r) ^nC_r, r = 0, 1, 2, ......, n$

$\therefore$ Sum of the series is given by

$\displaystyle S=\sum_{r=0}^n(-1)^r(3+5r)^nC_r$

$=3\displaystyle \sum_{r=0}^n(-1)^r {\;}^nC_r+5\displaystyle \sum_{r=0}^n(-1)^r r^nC_r$

$=3(\displaystyle \sum_{r=0}^n(-1)^r {\;}^nC_r)+5(\displaystyle \sum_{r=1}^n(-1)^r r.\frac {n}{r} ^{n-1}C_{r-1})$

$=3(\displaystyle \sum_{r=0}^n(-1)^r {\;}^nC_r)+5n (\displaystyle \sum_{r=1}^n(-1)^r {\;}^{n-1}C_{r-1})$

$=3(0) + 5n(0) = 0$

The sum of the coefficients in the expansion of

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

0%
$2^{10}$
0%
$4^{10}$
0%
$3^{10}$
0%
$1$

Explanation

Given expression is

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

Substituting

$x=y=z=1$ , we get the sum of the coefficients as

$(1+2+1)^{10}$

$=4^{10}$

If the

$9^{th}$ term in the expansion of

$\left(\displaystyle \frac{1}{x^2} + \frac{x}{2} log_2 x \right) ^{12}$ is equal to 495, then the value of x can be

0%
1
0%
An integer greater than 1
0%
a fraction
0%
an irrational number

Explanation

From the given question,

$\displaystyle T_9 = ^{12} C_4 \frac{1}{x^8} \frac{x^8}{2^8} (\log_2 x)^8 = 495$

$\Rightarrow \displaystyle \frac{12 \times 1 \times 10 \times 4}{24} \times \frac{1}{2^8} (\log_2 x)^8 = 495$

$\Rightarrow \log_2 x = \pm 2 \Rightarrow x = \dfrac{1}{4}$ or

$4$ .

In the expansion of

$\displaystyle \left [ 7^{1/3}+11^{1/9} \right ]^{6561}$ , the number of terms free from radicals is

0%
$730$
0%
$729$
0%
$725$
0%
$750$

Explanation

The general term of a binomial expansion is

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

From the above question

${ T }_{ r+1 }={ _{ }^{ 6561 }{ C } }_{ r }{ \left( { 7 }^{ \cfrac { 1 }{ 3 } } \right) }^{ 6561-r }\times { \left( { 11 }^{ \frac { 1 }{ 9 } } \right) }^{ r }={ _{ }^{ 6561 }{ C } }_{ r }{ 7 }^{ \frac { 6561 }{ 3 } -\frac { r }{ 3 } }{ 11 }^{ \frac { r }{ 9 } }.$

Number of terms free from radicals are those which makes the general term an integer

${ T }_{ r+1 }$ will be integer when

$\dfrac { 6561-r }{ 3 }$ and

$\dfrac { r }{ 9 }$ are positive integers for

$0\le r\le 6561$ .

Take

$r=0,{ \dfrac { 6561-0 }{ 3 } =2187 ,\dfrac { 6561 }{ 9 } =729 }$ is an integer.

For

$r=9$ ,

${ T }_{ r+1 }$ is an integer

Thus,

$r=0,9,18,27,..6561$ are the multiples of

$9$ , which makes

${ T }_{ r+1 }$ is an integer.

Hence,

$\dfrac { 6561 }{ 9 } =729$ There are

$729$ terms excluding

$0$

$\therefore$ Number of terms with free radicals

$=729+1=730$ terms (including the term 0)

In the expression of

$\displaystyle \left ( 3-\sqrt{\frac{17}{4}+3\sqrt{2}} \right )^{15}$ , the 11th term is a

0%
positive integer
0%
positive irrational number
0%
negative integer
0%
negative irrational number

Explanation

General term

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

$n=15$ ,

$x=3$

$a=\sqrt { \frac { 17 }{ 4 } +3\sqrt { 2 } } \\ { T }_{ 11 }={ T }_{ 10+1 }={ _{ }^{ 15 }{ C } }_{ 10 }{ (3) }^{ 15-10 }{ \left( \frac { 17 }{ 4 } +3\sqrt { 2 } \right) }^{ \frac { 1 }{ 2 } \times 10 }\\ ={ _{ }^{ 15 }{ C } }_{ 10 }{ 3 }^{ 5 }\left( \frac { 17 }{ 4 } +3\sqrt { 2 } \right) ^{ 5 }$

Note:

${ (3\sqrt { 2 } ) }^{ 5 }=972\sqrt { 2 }$ is an irrational number.

Sum of a rational number and an irrational number is irrational number.

Hence,11th term is a positive irrational number.

The total number of distinct terms in the expansion of

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

0%
$50$
0%
$202$
0%
$51$
0%
$none\ of\ these$

Explanation

Number of terms in a binomial expansion is

$=(n+1)$

When n is even,in the expansion

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

Thus,the odd number of terms get cancelled and even number of terms get added except the first term.

Therefore,total number of terms is=

$\left( \frac { n }{ 2 } +1 \right)$ terms

Hence, ${ (x+a) }^{ 100 }+{ (x-a) }^{ 100 }$ has

$\left( \frac { 100 }{ 2 } +1 \right) terms=51$ terms therefore Number of terms $=51$ terms

$C_0 , C_1 , C_2 .... C_n$ denote the binomial coefficients in the expansion of

$(1 + x)^n$ , then the value of

$\displaystyle \sum_{r=0}^n(r+1)C_r$ is

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

Explanation

$\sum(r+1)C_{r}$

$=\sum(r)C_{r}+\sum C_{r}$

$=\dfrac{d(1+x)^{n}}{dx}_{x=1}+2^{n}$

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

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

$(1+x-2x^2)^8= a_0 + a_1x + a_2x^2 +....+ a_{16}x^{16}$ then the sum

$a_1+a_3+a_5+ .... +a_{15}$ is equal to

0%
$-2^7$
0%
$2^7$
0%
$2^8$
0%
none of these

Explanation

Substituting x=1, we get

$0=a_{0}+a_{1}+a_{2}+...a_{16}$
Substituting x=-1 we get

$2^{8}=a_{0}-a_{1}+a_{2}+...-a_{15}+a_{16}$
Subtracting ii from i, we get

$-2^{8}=2[a_{1}+a_{3}+a_{5}+a_{7}+...a_{15}]$
Or

$a_{1}+a_{3}+a_{5}+a_{7}+...a_{15}=-2^{7}$

The number of terms which are free from radical signs in the expansion of

$(y^{\frac 15} + x^{\frac 1{10}})^{55}$ are

0%
5
0%
6
0%
7
0%
none of these

Explanation

In the expansion of

$(y^{1/5} + x^{1/10})^{55}$ ,the general term is

$T_{r+1} = ^{55}C_r (y^{1/5})^{55-r} (x^{1/10})^r = ^{55}C_r.y^{11-r/5}x^{r/10}$
This

$T_{r+1}$ will be independent of radicals if the exponents r/5 and r/10 are integers, for

$0\leq r \leq 55$ which is possible only when

$r = 0, 10, 20, 30, 40, 50$ .

$\therefore$ There are six terms viz.

$T_1,T_{11},T_{21},T_{31},T_{41},T_{51}$ , which are independent of radicals.

$(2x+3x^2)^6 = a_0 + a_1x + a_2x^2+.....+ a_{12} x^{12}$ , then values of

$a_0$ and

$a_6$ are

0%
$0, 6$
0%
$0, 2^6$
0%
$1, 6$
0%
$0$

Explanation

Writing the general term we get

$T_{r+1}$

$=\:^{6}C_{r}x^{6+r}2^{6-r}.3^{r}$

Hence coefficient of

$x^{0}$ is 0.

Thus

$a_{0}=0$ .
For

$a_{6}$ r=0.

Hence

$\:^{6}C_{0}.2^{6}.3^{0}$

$=2^{6}$ .

Hence

$a_{6}=2^{6}$ .

If sum of all the coefficients in the expansion of

$(x^{3/2} + x^{1/3})^n$ is 128, then the coefficient of

$x^5$ is

0%
35
0%
45
0%
7
0%
none of these

Explanation

Substituting

$x=1$ , we get the sum of the coefficients as

$(2)^{n}=128$

$\therefore n=7$
Hence writing the general term, we get

$T_{r+1}=\:^{7}C_{r}x^{\frac{63-11r}{6}}$
Hence for the coefficient of

$x^{5}$

$63-11r=6(5)$

$63-11r=30$

$33-11r=0$

$\therefore r=3$
Hence coefficient is

$\:^{7}C_{3}$

$=35$ .

If the second term of the expansion

$\displaystyle \left [ a^{1/13}+\frac{a}{\sqrt{a^{-1}}} \right ]^{n}\: \: is\: \: 14a^{5/2}$ , then the value of

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

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

Explanation

Using the formula

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

in the above question,we have

${ T }_{ 2 }={ _{ }^{ n }{ C } }_{ 1 }{ \left( { a }^{ \frac { 1 }{ 13 } } \right) }^{ n-1 }(a\sqrt { a } )=14{ a }^{ \frac { 5 }{ 2 } }$

$\Rightarrow { T }_{ 2 }={ _{ }^{ n }{ C } }_{ 1 }{ a }^{ \tfrac { n-1 }{ 13 } }{ a }^{ 1+\frac { 1 }{ 2 } }=14{ a }^{ \tfrac { 5 }{ 2 } }$

$\Rightarrow { _{ }^{ n }{ C } }_{ 1 }{ a }^{ \frac { n-1 }{ 13 } +\frac { 3 }{ 2 } }=14{ a }^{ \frac { 5 }{ 2 } }$

Comparing L.H.S and R.H.S,

$\Rightarrow { a }^{ \tfrac { 2n+37 }{ 26 } }={ a }^{ \tfrac { 5 }{ 2 } }$

Since bases are same,we can equate the powers

Thus,

$\cfrac { 2n+37 }{ 26 } =\cfrac { 5 }{ 2 }$

On solving ,

$2(2n+37)=5\times 26\Rightarrow 2n+37=65$

$\therefore 2n=65-37=28$

Hence,

$n=\cfrac { 28 }{ 2 } =14$

Using the formula,

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

$r=3$ and

$n=14$

$\cfrac { { _{ }^{ n }{ C } }_{ 3 } }{ { _{ }^{ n }{ C } }_{ 2 } } =\cfrac { 14-3+1 }{ 3 } =\cfrac { 12 }{ 3 } =4$

The number of real negative terms in the bionomial expansion of

$(1+ix)^{4n-2} ,n \in N,x>0$ is

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

Explanation

${ \left( 1+ix \right) }^{ 4x-2 }=\sum _{ r=0 }^{ ^{ 4x-2 } }{ ^{ ^{ 4x-2 } }{ { C }_{ r } } } { \left( ix \right) }^{ r }$

For real negative terms, we have to get

$i^2;$ Since

$(i^2=-1).$ For that,

$r$ must be equal to

$4k-2,$ where

$k\in I$

Hence, the total real negative terms in the expansion

$=n$

$p.s.$

$i=\sqrt { -1 }$

$\Rightarrow$

$i^2=-1$

$\Rightarrow$

$i^3=-\sqrt { -1 } =-i$

$\Rightarrow$

$i^4={ \left( { i }^{ 2 } \right) }^{ 2 }=(-1)^2=1$

Hence, the answer is

$n.$

Find

$7^{th}$ term of

$\displaystyle \left ( \frac{4x}{5}-\frac{5}{2x} \right )^{9}$

0%
$\dfrac{10050}{x^{3}}$
0%
$\dfrac{10500}{x^{3}}$
0%
$\dfrac{1050}{x^{3}}$
0%
$\dfrac{1000}{x^{3}}$

Explanation

$7^{th}$ term of

$\displaystyle \left ( \frac{4x}{5}-\frac{5}{2x} \right )^{9}$

$\displaystyle T_{6+1}=^{9}\textrm{C}_{6}\left ( \frac{4x}{5} \right )^{9-6}\left ( -\frac{5}{2x} \right )^{6}$

$\displaystyle =\frac{9!}{3!6!}\left ( \frac{4x}{5} \right )^{3}\left ( \frac{5}{2x} \right )^{6}=\frac{10500}{x^{3}}$

Find

$28^{th}$ term of

$\left ( 5x+8y \right )^{30}$

0%
$^{30}\textrm{C}_{27}\left ( 5x \right )^{27}\left ( 8y \right )^{3}$
0%
$^{30}\textrm{C}_{28}\left ( 5x \right )^{2}\left ( 8y \right )^{28}$
0%
$^{30}\textrm{C}_{27}\left ( 5x \right )^{3}\left ( 8y \right )^{27}$
0%
$^{30}\textrm{C}_{28}\left ( 5x \right )^{28}\left ( 8y \right )^{2}$

Explanation

$\displaystyle T_{27+1}=^{30}\textrm{C}_{27}\left ( 5x \right )^{30-27}\left ( 8y \right )^{27}=\frac{30!}{3!27!}\left ( 5x \right )^{3}.\left ( 8y \right )^{27}$

$p+q=1$ , then the value of

$\displaystyle \sum _{ r=0 }^{ 15 }{ { _{ }^{ 15 }{ C } }_{ r } } { p }^{ 15-r }{ q }^{ r }$

0%
$1$
0%
$15$
0%
$18$
0%
$20$

Explanation

It is given that

$p+q=1$

Therefore

$q=1-p$

Hence

$\sum_{r=0} ^{15} \:^{15}C_{r}p^{15-r}q^{r}$

$\quad =(p+q)^{15}$

Now

$q=1-p$

Hence

$(p+q)^{15}$

$=(p+1-p)^{15}$

$=1$ .

The value of

$\sum _{ r=1 }^{ 10 }{ { r }^{ 2 }.\cfrac { { _{ }^{ 26 }{ C } }_{ r } }{ { _{ }^{ 26 }{ C } }_{ r-1 } } }$ is-

0%
$1100$
0%
$1300$
0%
$900$
0%
$1800$

Explanation

$\sum _{ r=1 }^{ 10 }{ { r }^{ 2 } } .\left( \cfrac { 27-r }{ r } \right) =\sum _{ r=1 }^{ 10 }{ \left( 27r-{ r }^{ 2 } \right) }$

$=\cfrac { 27r(r+1) }{ 2 } -\cfrac { r(r+1).(2r+1) }{ 6 }$

put

$r=10$

$=27.5.11-5.11.7=1100$

${ \left( { x }^{ 2 }+\cfrac { 1 }{ { x }} \right) }^{ n }$ has exactly one middle term which is equal to

$\alpha.{x}^{3}$ then the value of

$(\alpha+n)$ is- (

$n\in N$ )

0%
$18$
0%
$21$
0%
$24$
0%
$26$

Explanation

$\because$ Only one middle term

$\Rightarrow$

$n$ is even middle term

$=\alpha.{x}^{3}$

${ _{ }^{ n }{ C } }_{ n/2 }{ \left( { x }^{ 2 } \right) }^{ n/2 }.{ \left( \cfrac { 1 }{ x } \right) }^{ n/2 }=\alpha .{ x }^{ 3 }$

${ _{ }^{ n }{ C } }_{ n/2 }x^{n/2}=\alpha .{ x }^{ 3 }\Rightarrow \cfrac { n }{ 2 } =3$
and

$\alpha ={ _{ }^{ n }{ C } }_{ n/2 }$

$\Rightarrow$

$n=6$ and

$\alpha={ _{ }^{ 6 }{ C } }_{ 3 }=20$

$\therefore \alpha +n=20+6=26$

The number of integral terms in

${(\sqrt {3}+\sqrt [ 8 ]{ 2 } )}^{64}$ is-

0%
$8$
0%
$7$
0%
$9$
0%
$6$

Explanation

The general term of expansion

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

${ ^{ n }{ C } }_{ r }{ x }^{ n-r }{ y }^{ r }$

So the general term of

${ (\sqrt { 3 } +\sqrt [ 8 ]{ 2 } ) }^{ 64 }$ is

${ ^{ 64 }{ C } }_{ r }{ 3 }^{ \frac { 64-r }{ 2 } }{ 2 }^{ \frac { r }{ 8 } }$

For the term to be integer ,

$r$ must be divided by

$8$ and

$64-r$ must be divided by

$2$

The possible values of

$r$ are

$0,8,16,24,32,40,48,56,64$

the number of integral values is

$9$

Value of

$\displaystyle \sum_{k=1}^{\infty} \sum _{r=0}^{k} \frac{1}{3^{k}} (^{k}C_{r})$ is

0%
$\displaystyle \frac{2}{3}$
0%
$\displaystyle \frac{4}{3}$
0%
$2$
0%
$1$

Explanation

$\sum _{ r=0 }^{ k }{ \dfrac { 1 }{ { 3 }^{ k } } } \left( { ^{ k }{ C } }_{ r } \right)$

$\Rightarrow \dfrac { 1 }{ { 3 }^{ k } } \sum _{ r=0 }^{ k }{ \left( { ^{ k }{ C } }_{ r } \right) } =\dfrac { 1 }{ { 3 }^{ k } }.2^k$

$={ \left( \dfrac { 2 }{ 3 } \right) }^{ k }$

$\sum _{ k=1 }^{ \infty }{ { \left( \dfrac { 2 }{ 3 } \right) }^{ k } }$

$=\dfrac{2}{3}+{ \left( \dfrac { 2 }{ 3 } \right) }^{ 2 }+{ \left( \dfrac { 2 }{ 3 } \right) }^{ 3 }........$

Sum of an infinite GP

$\Rightarrow S=\sum _{ k=1 }^{ \infty }{ { \left( \dfrac { 2 }{ 3 } \right) }^{ k } } =\dfrac { \dfrac { 2 }{ 3 } }{ 1-\dfrac { 2 }{ 3 } }=2$

Hence, the answer is

$2.$

Which term is the constant term in the expansion of

$\left ( 2x\, -\, \frac{7}{3x} \right )^6$ ?

0%
2nd term
0%
3rd term
0%
4th term
0%
5th term

If the middle term in the expansion of

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

$924x^6$ , then

$n=$

0%
$10$
0%
$12$
0%
$14$
0%
$none\ of\ these$

Explanation

For

$\left ( x^{2}+\dfrac{1}{x} \right )^{n}$ have middle term.

n is even

$\Rightarrow n = 2a; aEN$

middle term

$^{2a}C_{a} (x^{2})^{a}.\dfrac{1}{x^{a}} = ^{2a}C_{a}.x^{a}$

$= 924x^{6}$

$\Rightarrow a = 6$

Also for

$a = 6, ^{2a}C_{a} = ^{12}C_{6} = 924$

So,

$n = 2a = 12$

1170908_328582_ans_a07dc71baafa438d87e702277e1fa352.jpg

The middle term in the expansion of

$(x + 4)^4$ is:

0%
$96x^3$
0%
$96x^2$
0%
$-96x^2$
0%
none of the above

Explanation

Given

$(x+4)^4$

Here the middle terms can be calculated by

middle term

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

$=\dfrac{4}{2}+1$

$=3^{th} term$

$=\displaystyle^4C_{2}x^2 4^2$ $

$=\dfrac{4!}{2!2!}x^2\times 16$

$=\dfrac{4\times 3}{2\times 1}x^2\times 16$

$=96x^2$

Find the term which has the exponent of x as 8 in the expansion of

$\displaystyle\left(x^{\displaystyle\frac{5}{2}}-\frac{3}{x^3\sqrt{x}}\right)^{10}$

0%
$T_2$
0%
$T_3$
0%
$T_4$
0%
Does not exist

Explanation

Given,

$\left(x^{\dfrac{5}{2}}-\dfrac{3}{x^3\sqrt{x}}\right)^{10}$

General term,

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{10}C_r}(x^{\dfrac{5}{2}})^{10-r}\left(-\dfrac{3}{x^3\sqrt{x}}\right)^r$

$T_{r+1}={^{10}C_r}(x)^{25-\dfrac{5r}{2}}\dfrac{(-3)^r}{x^{\dfrac{7}{2}r}}$

$T_{r+1}={^{10}C_r}(-3)^r(x)^{25-6r}$

given the exponent of x as

$8$

$25-6r=8$

$25-8=6r$

$6r=17$

$r=\dfrac{17}{6}$

since r is a fraction number which is not possible. Hence, the term which has the exponents of x as

$8$ in the expansion of

$\left(x^{\dfrac{5}{2}}-\dfrac{3}{x^3\sqrt{x}}\right)^{10}$

does not exist.

1072451_397165_ans_0f44ba23094d4f2daab452aa97c2bcd2.png

Find the coefficient of the term independent of x in the expansion of

$\displaystyle\left(6x^3-\frac{5}{x^6}\right)^{12}$ .

0%
$^{12}C_45^86^4$
0%
$^{12}C_55^76^5$
0%
$^{12}C_46^85^4$
0%
None of these

Explanation

Given,

$\left(6x^3-\dfrac{5}{x^6}\right)^{12}$

General term,

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

To find coefficient of the term independent of x

$T_{r+1}={^{12}C_r}(6x^3)^{12-x}\left(\dfrac{-5}{x^6}\right)^r$

$T_{r+1}={^{12}C_r}(6)^{12-r}(x)^{36-3r}\dfrac{(-5)^r}{x^{6r}}$

$T_{r+1}={^{12}C_r}(6)^{12-x}(-5)^r(x)^{36-9r}$ ......

$(1)$

Since, term is independent of x.

$36-9r=0$

$9r=36$

$r=4$

Putting

$r=4$ in

$(1)$

$T_{r+1}={^{12}C_4}(6)^{12-4}(-5)^4(x)^0$

$T_{r+1}={^{12}C_4}(6)^8(5)^4$

So, the coefficient of the term independent of x in the expansion of

$\left(6x^3-\dfrac{5}{x^6}\right)^{12}$ is

$^{12}C_46^85^4$ .

If the coefficients of 6th and 5th terms of expansion

$(1+x)^n$ are in the ratio 7:5, then find the value of n.

0%
$11$
0%
$12$
0%
$10$
0%
$9$

Explanation

Given,

$(1+x)^n$

General term

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{n}C_r}(1)^{n-r}x^r$

Given

$\dfrac{\ coeff \ of\ T_6}{coeff\ of\ T_5}=\dfrac{7}{5}$

$T_6=T_{5+1}={^{n}C_5}(1)^{n-5}x^5$

$T_5=T_{4+1}={^{n}C_4}(1)^{n-4}x^4$

$\dfrac{^{n}C_5}{^{n}C_4}=\dfrac{7}{5}$

$\dfrac{n!4!\times (n-4)!}{5!\times (n-5)!\times n!}=\dfrac{7}{5}$

$\dfrac{n!\times 4!\times (n-4)\times (n-5)!}{5!\times (n-5)!\times n!}=\dfrac{7}{5}$

$\dfrac{(n-4)}{5}=\dfrac{7}{5}$

$n=7+4$

$n=11$

Hence, correct option is A.

If 'p' and 'q' are the coefficients of

$x^a$ and

$x^b$ respectively in

$(1+x)^{a+b}$ , then

0%
$2p=q$
0%
$p+q=0$
0%
$p=q$
0%
$p=2q$

Explanation

Given

$(1+x)^{a+b}$

General term

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{a+b}C_r}(1)^{a+b-r}(x)^r$

given 'p' the coefficeint of

$x^a$

$r=a$

$T_{r+1}={^{a+b}C_a}(1)^{a+b-a}(x)^a$

$T_{r+1}={^{a+b}C_a}(1)^b(x)^a$

given 'q' the coefficient of

$x^b$

$r=b$

$T_{r+1}={^{a+b}C_b}(1)^{a+b-b}(x)^b$

$T_{r+1}={^{a+b}C_b}(1)^a(x)^b$

$p={^{a+b}C_a}$

$q={^{a+b}C_b}={^{a+b}C_{a+b-b}}={^{a+b}C_a}$

Using

${^{n}C_r}={^{n}C_{n-r}}$

Hence

$p=q$ .

1072504_397169_ans_1010edab5dfb4e2aa218bcd08fe5ebb5.png

If sum of the first 3 coefficients is

$16$ in the expansion

$\displaystyle\left(x+\frac{1}{x^3}\right)^n$ , then find n.

0%
$10$
0%
$8$
0%
$5$
0%
$4$

Explanation

Given

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

General term.

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{n}C_r}(x)^{n-r}\left(\dfrac{1}{x^3}\right)^r$

$T_{r+1}={^{n}C_r}(x)^{n-r-3r}$

$T_{r+1}={^{n}C_r}(x)^{n-4r}$

Given, sum of the first three coefficients is

$16$

$T_1=T_{0+1}={^{n}C_0}(x)^n$

$T_2=T_{1+1}={^{n}C_1}(x)^{n-4}$

$T_3=T_{2+1}={^{n}C_2}(x)^{n-8}$

${^{n}C_0}+{^{n}C_1}+{^{n}C_2}=16$

$\dfrac{n!}{n!}+\dfrac{n!}{1!(n-1)!}+\dfrac{n!}{2!(n-2)!}=16\\$

$\dfrac{n(n-1)!}{(n-1)!}+\dfrac{n(n-1)(n-2)!}{2(n-2)!}=15\\$

$n+\dfrac{n(n-1)}{2}=15\\$

$\dfrac{2n+n^2-n}{2}=15\\$

$n^2+n=30\\$

$n^2+n-30=0\\$

$n^2+6n-5n-30=0\\$

$n(n+6)-5(n+6)=0\\$

$(n+6)(n-5)=0\\$

$\therefore$ Since n cannot be negative

so,

$n=5$

Hence, option C is correct.

$\displaystyle\sum_{r=2}^{16}{^{16}C_r}=$

0%
$2^{15}-15$
0%
$2^{16}-16$
0%
$2^{16}-17$
0%
$2^{17}-17$

Explanation

Consider given the binomial expression,

${{\sum\limits_{r=2}^{16}{^{16}{{C}_{r}}=}}^{16}}{{C}_{2}}{{+}^{16}}{{C}_{3}}{{+}^{16}}{{C}_{3}}+{{.......}^{16}}{{C}_{16}}$

$={{2}^{16}}-17$

Hence, this is the answer.

The value of the sum

${ \left( _{ }^{ n }{ { C }_{ 1 } } \right) }^{ 2 }+{ \left( _{ }^{ n }{ { C }_{ 2 } } \right) }^{ 2 }+{ \left( _{ }^{ n }{ { C }_{ 3 } } \right) }^{ 2 }+\dots +{ \left( _{ }^{ n }{ { C }_{ n } } \right) }^{ 2 }$ is

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

Explanation

We know that

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

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

On multiplying equations (i) and (ii), we get

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

Coefficient of

${ x }^{ n }$ in right hand side

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

and

coefficient of

${ x }^{ n }$ in left hand side

$=\ ^{ 2n }{ C }_{ n }$

$\therefore { \left( ^{ n }{ C }_{ 0 } \right) }^{ 2 }+{ \left(\ ^{ n }{ C }_{ 1 } \right) }^{ 2 }+\dots +{ \left( ^{ n }{ C }_{ n } \right) }^{ 2 }=\dfrac { 2n! }{ n!n! }$

$\Rightarrow { \left( ^{ n }{ C }_{ 1 } \right) }^{ 2 }+\dots +{ \left( ^{ n }{ C }_{ n } \right) }^{ 2 }=\dfrac { \left( 2n \right) ! }{ n!n! } -1$

$=\ ^{ 2n }{ C }_{ n }-1$

If the magnitude of the coefficient of

$\displaystyle { x }^{ 7 }$ in the expansion of

$\displaystyle { \left( { ax }^{ 2 }+\frac { 1 }{ bx } \right) }^{ 8 }$ , where a, b are positive numbers, is eual to the magnitude of the coefficient of

$\displaystyle { x }^{ -7 }$ in the expansion of

$\displaystyle { \left( { ax }^{ 2 }-\frac { 1 }{ bx } \right) }^{ 8 }$ , then a and b are connected by the relation

0%
$\displaystyle ab=1$
0%
$\displaystyle ab=2$
0%
$\displaystyle { a }^{ 2 }b=1$
0%
$\displaystyle { ab }^{ 2 }=2$

Explanation

Let the term

$\displaystyle { x }^{ 7 }$ in the expansion of

$\displaystyle { \left( { ax }^{ 2 }+\frac { 1 }{ bx } \right) }^{ 8 }is{ T }_{ r+1 }$

$\displaystyle \therefore \quad { T }_{ r+1 }=^{ 8 }{ C }_{ r }{ \left( { ax }^{ 2 } \right) }^{ 8-r }{ \left( \frac { 1 }{ bx } \right) }^{ r }$

$\displaystyle =^{ 8 }{ C }_{ r }\frac { { a }^{ 8-r } }{ { b }^{ r } } { x }^{ 16-3r }$
Since, this term contains

$\displaystyle { x }^{ 7 }$

$\displaystyle \therefore \quad 16-3r=7$

$\displaystyle \Rightarrow \quad 3r=9\Rightarrow r=3$

$\displaystyle \therefore$ Coefficient of

$\displaystyle { x }^{ 7 }$ in the expansion of

$\displaystyle { \left( { ax }^{ 2 }+\frac { 1 }{ bx } \right) }^{ 8 }=^{ 8 }{ C }_{ r }.\frac { { a }^{ 5 } }{ { b }^{ 3 } }$
Also, the term containing

$\displaystyle { x }^{ -7 }$ in the expansion of

$\displaystyle { \left( { ax }-\frac { 1 }{ bx } \right) }^{ 8 }is{ T }_{ R+1 }$

$\displaystyle { T }_{ R+1 }=^{ 8 }{ C }_{ r }{ \left( ax \right) }^{ 8-R }{ \left( -\frac { 1 }{ { bx }^{ 2 } } \right) }^{ R }$

$\displaystyle =^{ 8 }{ C }_{ r }\frac { { a }^{ 8-R }{ x }^{ 8-R }{ \left( -1 \right) }^{ R } }{ { b }^{ R }{ x }^{ 2R } }$

$\displaystyle ={ \left( -1 \right) }^{ R }\quad ^{ 8 }{ C }_{ r }\frac { { a }^{ 8-R } }{ { b }^{ R } } .{ x }^{ 8-3R }$
Since, this term contains

$\displaystyle { x }^{ -7 }$

$\displaystyle \therefore \quad 8-3R=-7$

$\displaystyle \Rightarrow 3R=15\Rightarrow \quad R=5$

$\displaystyle \therefore$ Coefficient of

$\displaystyle { x }^{ -7 }$ in the expansion of

$\displaystyle { \left( ax-\frac { 1 }{ { bx }^{ 2 } } \right) }^{ 8 }$
According to the given condition,

$\displaystyle \left| { T }_{ R+1 } \right| =\left| { T }_{ R+1 } \right|$

$\displaystyle \Rightarrow \quad ^{ 8 }{ C }_{ 3 }.\frac { { a }^{ 5 } }{ { b }^{ 3 } } =^{ 8 }{ C }_{ 5 }.\frac { { a }^{ 3 } }{ { b }^{ 5 } }$

$\displaystyle \Rightarrow \quad { a }^{ 2 }{ b }^{ 2 }=1$

$\displaystyle \Rightarrow \quad ab=1$

The value of

$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right )$ is equal to:

0%
$680$
0%
$1085$
0%
$560$
0%
$1240$

Explanation

$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right )$

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

$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right ) = \displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {n-r+1}{r}\right )\\$

$\sum_{r = 1}^{15} (nr-r^2+r) = \sum_{r = 1}^{15} ((n+1)r-r^2)$

$\sum_{r = 1}^{15} (nr-r^2+r) = \dfrac{n(n+1)^2}{2}- \dfrac{n(n+1)(2n+1)}{6}$

Put

$n=15,$ we get

$\displaystyle \sum_{r = 1}^{15} r^{2} \left (\dfrac {^{15}C_{r}}{^{15}C_{r - 1}}\right ) = 680$

The middle term in the expansion of

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

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

Explanation

Middle term in the expansion of

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

$= t_{n + 1}$

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

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

$= \dfrac {\overline {2n}(2n - 1)(\overline {2n - 2})(2n - 3) .... \overline {4}\times 3\times \overline {2} \times 1}{(n!)(n!)} . x^{n}$

$= \dfrac {2^{n}[n(n - 1)(n - 2) ....\times 2\times 1][(2n - 1)(2n - 3) .... 3\times 1]}{(n!)(n!)} x^{n}$

$=\dfrac {(2n - 1)(2n - 3) ....3\times 1}{n!} 2^{n}x^{n}$

The number of irrational terms in the binomial expansion of

$(3^{1/5}+7^{1/3})^{100}$ is

0%
$90$
0%
$88$
0%
$94$
0%
$95$

Explanation

General term of

$(3^{1/5}+7^{1/3})^{100}$ is given by,

$T_{r+1}=\ ^{100}C_r(3^{1/5})^{100-r}(7^{1/3})^r$

$\displaystyle =\ ^{100}C_r\cdot 3^{\frac{100-r}{5}}\cdot 7^{\frac{r}{3}}$

For a rational term,

$\dfrac{100-r}{5}$ and

$\dfrac{r}{3}$ must be integer.

Hence,

$r=0, 15, 30, 45, 60, 75, 90$

$\therefore$ There are seven rational terms.

Hence, number of irrational terms

$=101-7=94$

If in the expansion of

$(a-2b)^n$ , the sum of the

$5th$ and

$6th$ term is zero, then the value of

$\dfrac{a}{b}$ is

0%
$\dfrac{n-4}{5}$
0%
$\dfrac{2(n-4)}{5}$
0%
$\dfrac{5}{n-4}$
0%
$\dfrac{5}{2n-4}$

Explanation

We know,

$(a-2b)^n=\sum_{r=0}^{n}\,{^n{C_r}}(a)^{n-r}(-2b)^r$
The

$(r+1)$ th term

$=t_{r+1}={^n{C_r}}(a)^{n-r}.(-2b)^r$

$\therefore\;t_5+t_6=0$

$\Rightarrow\,{^n{C_4}}(a)^{n-4}(-2b)^4+{^n{C_5}}(a)^{n-5}(-2b)^5=0$

$\Rightarrow\;\dfrac{n!}{4!(n-4)!}a^{n-4}(-2b)^4$

$=-\dfrac{n!}{5!(n-5)!}(a)^{n-5}(-2b)^5$

$\Rightarrow\;\dfrac{1}{(n-4)}\times a=\dfrac{-1}{5}(-2b)$

$\Rightarrow\,\dfrac{a}{b}=\dfrac{2(n-4)}{5}$

The middle term in the expansion of

$\left (x - \dfrac {1}{x}\right )^{18}$ is

0%
$^{18}C_{9}$
0%
$-^{18}C_{9}$
0%
$^{18}C_{10}$
0%
$-^{18}C_{10}$

Explanation

The general term in the expansion

$\left (x - \dfrac {1}{x}\right )^{18}$ is given by

$T_{r + 1} = ^{18}C_{r} (x)^{18 - r} \left (-\dfrac {1}{x}\right )^{r}$
Here,

$n = 18$

$\therefore$ The middle term is

$T_{9 + 1}$ , where

$r = 9$

$\therefore T_{9 + 1} = ^{18}C_{9} (-1)^{9} x^{18 - 2r}$

$= -^{18}C_{9}x^{18 - 18} = -^{18}C_{9}$

${ \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 } \right) }^{ 5 }=\displaystyle\sum _{ k=0 }^{ 15 }{ { a }_{ k }{ x }^{ k } }$ then

$\displaystyle\sum _{ k=0 }^{ 7 }{ { a }_{ 2k } }$ is equal to

0%
$128$
0%
$256$
0%
$512$
0%
$1024$

Explanation

Given,

${ \left( 1+x+{ x }^{ 2 }+{ x }^{ 3 } \right) }^{ 5 }=\displaystyle\sum _{ k=0 }^{ 15 }{ { a }_{ k }{ x }^{ k } }$

$\Rightarrow { \left[ \left( 1+x \right) +x\left( 1+x \right) \right] }^{ 5 }=\displaystyle\sum _{ k=0 }^{ 15 }{ { a }_{ k }{ x }^{ k } }$

$\Rightarrow { \left( 1+x \right) }^{ 10 }={ a }_{ 0 }{ x }^{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+\dots +{ a }_{ 15 }{ x }^{ 15 }$

$\Rightarrow _{ }^{ 10 }{ { C }_{ 0 } }+_{ }^{ 10 }{ { C }_{ 1 } }x+_{ }^{ 10 }{ { C }_{ 2 } }{ x }^{ 2 }+\dots +_{ }^{ 10 }{ { C }_{ 10 } }{ x }^{ 10 }$

$={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 3 }{ x }^{ 3 }+\dots +{ a }_{ 15 }{ x }^{ 15 }$
On equating the coefficient of constant and even powers of

$x$ , we get

${ a }_{ 0 }=_{ }^{ 10 }{ { C }_{ 0 } },\quad { a }_{ 2 }=_{ }^{ 10 }{ { C }_{ 2 } },$

${ a }_{ 4 }=_{ }^{ 10 }{ { C }_{ 4 } },\dots ,{ a }_{ 10 }=_{ }^{ 10 }{ { C }_{ 10 } },$

${ a }_{ 12 }={ a }_{ 14 }=0$

$\therefore \displaystyle\sum _{ k=0 }^{ 7 }{ { a }_{ 2 }k } =_{ }^{ 10 }{ { C }_{ 0 } }+_{ }^{ 10 }{ { C }_{ 2 } }+^{ 10 }{ { C }_{ 4 } }+^{ 10 }{ { C }_{ 6 } }+^{ 10 }{ { C }_{ 8 } }+^{ 10 }{ { C }_{ 10 } }+0+0$

$={ 2 }^{ 10-1 }=29$

$=512$

Find the sum of coefficients in the expansion of

$\left(1-\dfrac 2x+\dfrac {4}{x^2}\right)^n$ given that the number of terms are

$28$

0%
$27$
0%
$81$
0%
$243$
0%
$729$

Explanation

$\left(1-\dfrac 2x+\dfrac {4}{x^2}\right)^n$ and number of terms

$=28$

Number of terms

$=^{n+3-1}C_{3-1} = ^{n+2}C_2$

$\dfrac {(n+2)!}{2! (n!)} = \dfrac {(n+2)(n+1)}{2} = 28$

$n^2+3n-54=0$

$n^2-6n+9n-54=0$

$(n+3)(n-6)=0$

$n=6$

Number of terms

$=\left(1-\dfrac 2x+\dfrac {4}{x^2}\right)^6$

Sum of coefficients is known when

$x=1$

No. of terms

$=(1-2+4)^6 = 3^6 = 729$

$(x+1)^n=C_0+C_1x^1+C_2x^2....C_nx^n$ .

Find the value of

$C_0+C_1+C_2.....+C_n$

0%
$n!$
0%
$(n!)^2$
0%
$2^n$
0%
$n!-2^n$

Explanation

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

Put

$x=1$ ,

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

Hence,

$C$ is correct.

Let

$n \in N$ , such that

$(1+x+x^2)^n=a_0+a_1x+a_2x^2...a_{2n}x^{2n}$ The value of

$a_r$ when

$(0\leq r\leq 2n)$

0%
$a_{2n-r}$
0%
$a_{n-r}$
0%
$a_{2n}$
0%
$n.a_{2n-1}$

Explanation

We have

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

$=$

$\sum _{r=0 }^{2n }{ a_rx^r}$

Replace

$x$ by

$\dfrac{1}{x}$

Therefore,

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

$\sum _{r=0 }^{2n }{ a_rx^r}$

$\Rightarrow$

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

$=$

$\sum _{r=0 }^{2n }{ a_r{x}^{2n-r}}$

$\Rightarrow$

$\sum _{r=0 }^{2n }{ a_rx^r}$

$=$

$\sum _{r=0 }^{2n }{ a_r{x}^{2n-r}}$

So, equating both the sides, we get

$a_r$

$=$

$a_{2n-r}$ .

Find the number of integral terms in the expansion of

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

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

Explanation

$(3^{1/2}+5^{1/8})^{256}$

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

For integral terms

$\dfrac{256-r}{2},\dfrac{r}{8}$ are both positive integer

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

$256=0+(n-1)8$ using

$t_n=a+(n-1)d$

$\dfrac{256}{8}=n-1$

$n=\dfrac{256}{8}+1$

$n$

$=32+1=33$

Find the coefficient of

$x^n$ in expansion of

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

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

The general term in

$(x + y)^n$ is given by

$T_{r+1}=$

0%
$^{n}C_{r+1}$
0%
$^nC_{n-r-1}$
0%
$^{n-1}C_r$
0%
none

Explanation

The general term

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

So,

$D$ is correct.

If the middle term of

$\left(\dfrac 1x +x\sin x\right)^{10}$ is equal to

$7\dfrac 78$ , then the principal value of

$x$ is:

0%
$60^0$
0%
$30^0$
0%
$45^0$
0%
$90^0$

Explanation

Solution:

Middle term of the expansion

$\left(\cfrac1x+x\sin x\right)^{10}$ is

$T_6.$

$T_6={}^{10}C_5\times\left(\cfrac1x\right)^5\times(x\sin x)^5=7\cfrac78$

or,

${}^{10}C_5\times \sin^5x=\cfrac{63}8$

or,

$252\times \sin^5x=\cfrac{63}8$

or,

$\sin^5x=\cfrac{1}{32}$

or,

$\sin x=\cfrac 12$

or,

$x=30^\circ$

Hence, B is the correct option.

If the value of

$C_0+2 \cdot C_1 + 3 \cdot C_2+........+(n+1)\cdot C_n=576$ , then n is ______.

0%
$7$
0%
$5$
0%
$6$
0%
$9$

Explanation

We have:

$(1 + x)^n = C_0 + C_1x + C_2x^2 + ... + C_nx^n$

Multiplying by

$x$ on both sides, we get

$x(1 + x)^n = C_0x + C_1x^2 + C_2x^3 + ... + C_nx^{n + 1}$

Differentiating both sides by

$x$ , we get

$xn(1 + x)^{n - 1} + (1 + x)^n = C_0 + 2C_1x + 3C_2x^2 + ... + (n + 1)C_nx^n$

Substituting

$x = 1$ , we have

$n2^{n - 1} + 2^n = C_0 + 2C_1 + 3C_2 + ... + (n + 1)C_n$

Thus, the R.H.S. is what has been asked, and so

$n2^{n - 1} + 2^n = 576$

i.e.

$2^{n - 1} (2 + n) = 576$

$n = 5, n + 2$ becomes

$7$ and

$576$ is not divisible by

$7$ .

Similarly,

$n$ cannot be

$9$ .

When

$n = 7, 2^6 \times 9 = 64 \times 9 = 576$

Ans is option

$A$ .

The value of

$a_0 ^2-a_1 ^2+a_2 ^2....a_{2n} ^2$ is

0%
$a_n$
0%
$2a_n$
0%
$3a_n$
0%
$\dfrac {n+1}2 (a_0+a_1+a_2...a_{2n})$

Explanation

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

or,

$(1+x)^{2n}=a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$ ...

$(i)$

Similarly,

$(1-x)^{2n}=a_0-a_1x+a_2x^2+...+a_{2n}x^{2n}$ ....

$(ii)$

As,

$a_0=a_{2n}$ ,

$a_1=a_{2n-1},...$ and so on. So, (i) and (ii) can be written as:

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

$(1+x)^{2n}=a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$

So, required answer is coefficient of

$x^{2n}$ in

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

$x^{2n}$ in

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

${T}_{r+1}=_{ }^{2n}{C}_r(-x^2)^r$

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

So, we need,

$r=n$ . Hence, the answer is

$(-1)^n.a_n$

For

$n-even$ , the answer is option

$A$ , i.e;

$a_n$

The first three terms in the expansion of $(1+a)^n$ are $t_1=1,t_2=-18,t_3=144$ .

Use the general term to determine a and n.

0%
$a=-3,n=9$
0%
$a=-2,n=9$
0%
$a=-3,n=8$
0%
$a=-2,n=8$

Explanation

As we can see from option that n is positive so we proceed with the expansion

$T_1$

$=$

$^{n}{C}_{0}$

$=1$

$T_2$

$=$

$^{n}{C}_{1}a$

$=n.a=-18$ ....(1)

$T_3$

$=$

$^{n}{C}_{2}a^2$

$=$

$\dfrac{(n)(n-1)}{2}$ .

$a^2$

$=144$ ....(2)

$\Rightarrow$

${(na)(na-a)}$ .

$=288$

Putting the value of

$na$ from (1), we get

$(-18)(-18-a)$

$=288$

$\Rightarrow$

$-18-a$

$= -16$

$\Rightarrow$

$-a=2$

So,

$a= -2$

Substitute this value of

$a$ in equation (1), we get

$na=-18$

$\Rightarrow n(-2)=-18$

$\Rightarrow n=9$

Thus value of

$a$ and

$n$ are

$-2$ and

$9$ .

The middle term of expansion of

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

0%
$^{7}C_{5}$
0%
$^{8}C_{5}$
0%
$^{9}C_{5}$
0%
$^{10}C_{5}$

Explanation

Given

Fact: middle in the expansion of

$(x+y)^n$ will be

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

$n$ is even

Hence

$n=10=$ even

So middle term will be

$6$ th term

Which is given by,

$^{10}C_5\left(\dfrac{10}{x}\right)^5\left(\dfrac{x}{10}\right)^5=^{10}C_5$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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