Assertion (A) : Number of the disimilar terms in the sum of expansion $$(x+a)^{102}+(x-a)^{102}$$ is $$206$$ Reason (R) : Number of terms in the expansion of $$(x+b)^{n}$$ is n + 1

Both A and R are individually true and R is the correct explanation of A.

Both A and R are individually true and R is not correct explanation of A.

MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 2

The coefficient of

$\displaystyle \frac{1}{{x}}$ in the expansion of

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

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$\displaystyle \frac{\mathrm{n}!}{(\mathrm{n}-1)!(\mathrm{n}+1)!}$
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$\displaystyle \frac{2\mathrm{n}!}{(\mathrm{n}-1)!(\mathrm{n}+1)!}$
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$\displaystyle \frac{\mathrm{n}!}{(2\mathrm{n}-1)!(2\mathrm{n}+1)!}$
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2n!(2n−1)!(2n+1)!

Explanation

Given,

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

$=\dfrac{(1+x)^{2n}}{x^{n}}$ ...(i)
Therefore

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

$=\:^{2n}C_{r}x^{r-n}$ ...(ii)
For coefficient of

$x^{-1}$

$r-n=-1$

$r=n-1$
Substituting in (ii), we get

$T_{n-2}=\:^{2n}C_{n-1}x^{-1}$

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

Coefficient of

$x$ in the expansion of

$(1-2\displaystyle {x}^{3}+3{x}^{5})(1+\frac{1}{{x}})^{8}$ is

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$154$
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$164$
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$146$
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$156$

Explanation

$(1-2x^3+3x^5)(1+x^{-1})^{8}$

$=(1-2x^3+3x^5)(1+{}^8C_1x^{-1}+\:^{8}C_{2}x^{-2}+\:^{8}C_{3}x^{-3}...+\:^{8}C_{8}x^{-8})$ .....

$(1)$

Hence coefficient of

$x$ will be

$-2\times \:^{8}C_{2}+3\times \:^{8}C_{4}$ since after solving (1) we get

$x$ at

$x^{3-2}$ and

$x^{5-4}$

$=-2\times \dfrac{8.7}{2}+3\times \dfrac{8.7.6.5}{4!}$

$=-56+\dfrac{8.7.6.5}{2.4}$

$=210-56$

$=154$

$a$ is the coefficient of the middle term in the expansion of

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

$b, c$ are the coefficients of the two middle terms in the expansion of

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

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$a +b = c$
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$a = b + c$
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$a=b=c$
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$b = a + c$

Explanation

For

$(1+x)\:^{2n}$
Middle term

$=\dfrac{N}{2}+1$ since 2n is even

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

$=(n+1)^{th}$ term
Now consider the following

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(i)
Where

$T$ represents the term
Here the middle term is the

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

$r+1=n+1$

$r=n$
Substituting in (i), we get

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

$=\:^{2n}C_{n}x^{n}$
Therefore

$a=\:^{2n}C_{n}$ because a is the coefficient ...(a)

Consider

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

$2n-1$ is odd the middle terms are

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

$(\dfrac{N+1}{2}+1)$ th terms
Now consider the following

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(ii)
Where

$T$ represents the term
Here the middle terms are the

$(n)^{th}$ term and

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

$r+1=n+1$ and

$r+1=n$

$r=n$ and

$r=n-1$
For

$r=n$

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

$=\:^{2n-1}C_{n}x^{n}$
Hence,

$b=\:^{2n-1}C_{n}$ , since

$b$ is the coefficient ...(b)

Similarly for

$r=n-1$

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

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

$c=\:^{2n-1}C_{n-1}$ since

$c$ is the coefficient ...(c)
From

$a,b,c$ , we get

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

$a=b+c$

The middle term in the expansion of

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

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$6{n}_{{C}_{3{n}}}(-{x})^{3{n}}$
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$6{n}_{{C}_{2{n}}}(-{x})^{2{n}+1}$
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$4{n}_{{C}_{3{n}}}(-{x})^{3{n}}$
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$6{n}_{{C}_{3{n}}}(-{x})^{3{n}-1}$

Explanation

Given,

$(1-3x+3x^2-x^3)\:^{2n}$

$=[(1-x)\:^3]\:^{2n}$

$=(1-x)\:^{6n}$

Middle term

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

$=\dfrac{6n}{2}+1$

$=(3n+1)^{th}$ term

Now consider the following,

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(i)

Where

$T$ represents the term

Here the middle term is the

$(3n+1)^{th}$ term.

Hence,

$r+1=3n+1$

$r=3n$

Substituting in (i), we get

$(-1)\:^{3n}\:^{6n}C_{3n}1^{6n-3n}x^{3n}$

$=(-1)\:^{3n}\:^{6n}C_{3n}x^{3n}$

$=\:^{6n}C_{3n}(-x)\:^{3n}$

Hence answer is A

Sum of the coefficients in the expansion of

$(5x-4y)^n$ where

$n$ is a positive integer is

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$1$
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$9^n$
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$(-1)^n$
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$5^n$

Explanation

Substituting

$x=1$ and

$y=1$ in the above expression we get the sum of the binomial coefficients as

$(5x-4y)^{n}$

$=(5-4)^{n}$ (substituting x=1 and y=1)

$=(1)^{n}$

$=1$
Hence, answer is

$A.$

$n$ is a positive integer, then the coefficient of

$x^n$ in the expansion of

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

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$n.3^n$
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$(n-1)3^n$
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$(n+1)3^n$
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$3^n$

Explanation

To find the coefficient of

$x^{n}$ in the expansion of

$\dfrac{(1+x)^{2n}}{1-x}$
This can be written as

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

$= \left ( \binom{2n}{0} + \binom{2n}{1}x + \binom{2n}{2}x^{2} + \binom{2n}{3}x^{3}+............+ \binom{2n}{n} x^{n} \right ) \left ( 1+ x+x^{2}+ x^{3}+x^{4}+.. . \right )$

On Multiplying this, we can find the coefficient of

$x^{n}$ as summation of

$= \binom{2n}{0} + \binom{2n}{1} + \binom{2n}{2} .......+\binom{2n}{n}$

$= 3^{n}$

The coefficient of

${x}^{p}$ the expansion of

$(\displaystyle {x}^{2}+\frac{1}{{x}})^{{n}}$ , when it exists is

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$2n_{{C}_{\frac{4{n}+{p}}{3}}}$
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$2n_{{C}_{\frac{2{n}+{p}}{3}}}$
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$2n_{{C}_{\frac{{n}+{p}}{3}}}$
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$n_{{C}_{\frac{{n}+{p}}{3}}}$

Explanation

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

$T_{r+1}=\:^{2n}C_{r}x^{4n-3r}$ ...(i)
For the coefficient of

$x^p$

$4n-3r=p$

$\dfrac{4n-p}{3}=r$
Substituting in equation (i), we get

$T_{(\frac{4n-p}{3})+1}=\:^{2n}C_{\frac{4n-p}{3}}x^{p}$
Hence, coefficient of

$x^p$ in the above binomial expansion is

$\:^{2n}C_{\frac{4n-p}{3}}$ .

$n$ is a positive integer, then the coefficient of

$x^n$ in the expansion of

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

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

Explanation

To find the coefficient of

$x^{n}$ in the expansion of

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

This can be written as

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

$= \left ( \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^{2} + \binom{n}{3}x^{3}+............+ \binom{n}{n} x^{n} \right ) \left ( 1-x+x^{2}-x^{3}+x^{4}-......... \right )$

On Multiplying this, we can find the coefficient of

$x^{n}$ as

Summation of

$= C_{0} + C_{1}+ C_{2}+ ............ C_{n}$

$= 2^{n}$

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

$x$ of

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

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

$\alpha=$

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-5/3
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3/5
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-3/10
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10/3

Explanation

The middle term for

$(1+\alpha x)\:^4$ is

$3^{rd}$ term and for

$(1-\alpha x)\:^6$ is

$4^{th}$ term.

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(i)
Where

$T$ represents the term
Applying formula (i) to find the

$3^{rd}$ term of

$(1+\alpha x)\:^4$ and to find

$4^{th}$ term of

$(1-\alpha x)\:^6$ , we get

$3^{rd}$ term of

$(1+\alpha x)\:^4$ is

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

$4^{th}$ term of

$(1-\alpha x)\:^6$ is

$(-\:^{6}C_{3}(\alpha x)\:^{3})$
Applying the condition given in the question, we get

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

$\dfrac{4!}{2!\:2!}=-\dfrac{6!}{3!\:3!}\alpha$

$(3.2)=-(4.5)\alpha$

$\alpha=-\dfrac{6}{20}$

$\alpha=-\dfrac{3}{10}$
Hence answer is C

The sum of the coefficients in the expansion of

$(1-x)^{10}$

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$0$
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$1$
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$-1$
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$2^{10}$

Explanation

$(1-x)^{10}$

$=1-\:^{10}C_{1}x+\:^{10}C_{2}x^2+...\:^{10}C_{10}x^{10}$
Substituting

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

$1-\:^{10}C_{1}+\:^{10}C_{2}+...\:^{10}C_{10}$

$=(1-1)^{10}$

$=0$

In the expansion of

$(1+{x})^{\mathrm{n}}$ , the

$5^{{t}{h}}$ term is

$4$ times the

$4^{{t}{h}}$ term and the

$4^{{t}{h}}$ term is

$6$ times the

$3^{{r}{d}}$ term. than

$n =$

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9
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10
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11
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12

Explanation

It is given in the question that

$T_{5}=4T_{4}$

$\:^nC_{4}x^4=4\:^nC_{3}x^3$

$\:^nC_{4}x=4\:^nC_{3}$

$\dfrac{x}{4}=\dfrac{4}{n-3}$

$x=\frac{16}{n-3}$ ...(i)
And

$T_{4}=6T_{3}$

$\:^nC_{3}x^3=6\:^nC_{2}x^2$

$\dfrac{x}{3}=\dfrac{6}{n-2}$

$x(n-2)=18$
Substituting the value of

$x$ from Eq (i), we get.

$\dfrac{16(n-2)}{n-3}=18$

$8n-16=9n-27$

$n=27-16$

$n=11$

Assertion (A) : The coefficient of

$x^{7}$ in

$(\displaystyle \frac{x^{2}}{2}-\frac{2}{x})^{9}$ is zero

Reason (R) :

$r$ in

$t_{r+1}$ that contains coefficient of

$x^{7}$ is not positive integer

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Both A and R are true and R is the correct explanation of A
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Both A and R are true and R is not the correct explanation of A
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A is true, but R is false
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A is false, but R is true

Explanation

$(\dfrac{x^2}{2}-2x^{-1})^{9}$

$T_{r+1}=(-1)^{r}\:^9C_{r}x^{18-3r}2^{2r-9}$
For the coefficient of

$x^{7}$ ,

$18-3r=7$

$3r=11$

$r=\frac{11}{3}$
Since, the value of

$r$ for the coefficient of

$x^7$ is not a positive integer. Hence, the coefficient of

$x^7$ is zero.

lf the

$5^{\mathrm{t}\mathrm{h}}$ term of

$(\displaystyle \frac{\mathrm{q}}{2\mathrm{x}}-\mathrm{p}\mathrm{x})^{8}$ is 1120 and

$\mathrm{p}+\mathrm{q}=5,\ \mathrm{p}>\mathrm{q}$ then

$\mathrm{p}=$

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3
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6
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4
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7

Explanation

$T_{5} = 1120$

$1120=\:^nC_{4}\frac{q^4}{16x^{4}}p^{4}x^{4}$

$1120=\frac{70.(p.q)^{4}}{16}$

$\displaystyle \frac{70.(p.q)^{4}}{16}=1120$

$\displaystyle \frac{(p.q)^{4}}{16}=16$

$(p.q)^{4}=16^{2}$

$(p.q)^{4}=4^4$

$p.q=4$
It is given that

$p+q=5$
Substituting

$q=\frac{4}{p}$ in

$p+q=5$ we get

$p^2+4=5p$

$p^2-5p+4=0$

$(p-1)(p-4)=0$

$p=4$ or

$p=1$

$(1+\mathrm{k}{\mathrm{x}})^{10}=\mathrm{a}_{0}+\mathrm{a}_{1}\mathrm{x}+\mathrm{a}_{2}\mathrm{x}^{2}+\ldots.+\mathrm{a}_{10}\mathrm{x}^{10}$ and

$\displaystyle \mathrm{a}_{2}+\frac{7}{5}\mathrm{a}_{1}+1=0$ then

$\mathrm{k}=$

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-1/5, -1/9
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1/5, 1/9
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-1/5, -1/7
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1/5, -1/9

Explanation

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$
Applying to the above question we get

$a_{1}=\:^{10}C_{1}k$

$=10k...(i)$

$a_{2}=\:^{10}C_{2}k^2$

$=45k^2$ ...(i)
It is given that

$a_{2}+\displaystyle \frac{7}{5}a_{1}+1=0$ ..(iii)
Substituting the value of

$a_{1}$ and

$a_{2}$ in (iii) we get

$14k+45k^2+1=0$

$45k^2+14k+1=0$

$(9k+1)(5k+1)=0$
Hence

$k=\displaystyle \frac{-1}{5},\displaystyle \frac{-1}{9}$

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

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

Explanation

As in the hint required expression

$+$

$^{n+1}C_0 + ^{n+1}C_{n+1} = 2^{n+1}$
=> required. expression

$=$

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

The ratio of

$(r + 1) ^{th}$ and

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

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

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$\displaystyle \frac{(n-r+2)(n-r+1)}{r(r-1)}\cdot\frac{b^{2}}{\mathrm{a}^{2}}$
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$\displaystyle \frac{(n-r+2)(n-r+1)}{r(r-1)}.\frac{\mathrm{a}^{2}}{b^{2}}$
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$(\displaystyle \frac{n-r+2}{r})\cdot\frac{b}{\mathrm{a}}$
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$(\displaystyle \frac{n-r+1}{r-1})\cdot\frac{b}{\mathrm{a}}$

Explanation

We know,

$T_{r+1}=\:^nC_{r}a^{n-r}b^{r}$
Therefore

$\displaystyle \frac{T_{r+1}}{T_{r-1}}$

$=\displaystyle \frac{\:^nC_{r}a^{n-r}b^{r}}{\:^nC_{r-2}a^{n-r+2}b^{r-2}}$

$=\displaystyle \frac{\:^nC_{r}b^2}{\:^nC_{r-2}a^2}$

$=\displaystyle \frac{n!(n-(r-2))!(r-2)!b^2}{n!(n-r)!r!a^2}$

$=\displaystyle \frac{(n-(r-2))(n-(r-1))b^2}{r(r-1)a^2}$

$=\displaystyle \frac{(n-r+2)(n-r+1)b^2}{r(r-1)a^2}$

If the

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

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

$(a+b)^n$ are

$135, 30$ and

$\dfrac {10}{3}$ respectively, then the value of

$n$ is:

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$5$
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$6$
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$7$
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$8$

Explanation

It is given that

$T_{2}=na^{n-1}b=135$

$T_{3}=\displaystyle \frac{n(n-1)a^{n-2}b^2}{2}=30$

$T_{4}=\displaystyle \frac{n.(n-1)(n-2)a^{n-3}b^{3}}{6}=\frac{10}{3}$
Therefore, by taking ratios, we get

$\Rightarrow \displaystyle \frac{na^{n-1}b}{\frac{n(n-1)a^{n-2}b^2}{2}}=\frac{135}{30}$

$\Rightarrow \displaystyle \frac{2a}{(n-1)b}=\frac{9}{2}$

$\Rightarrow 4a=9b(n-1)$ ...(i)

$\Rightarrow \displaystyle \frac{\frac{n(n-1)a^{n-2}b^2}{2}}{\frac{n.(n-1)(n-2)a^{n-3}b^{3}}{6}}=\frac{3(30)}{10}$

$\Rightarrow \displaystyle \frac{3a}{b(n-2)}=9$

$\Rightarrow a=3b(n-2)$

$\Rightarrow \displaystyle \frac{a}{b}=3(n-2)$ ...(ii)

Substituting the value of

$\dfrac{a}{b}$ in equation (i), we get

$\Rightarrow \displaystyle \frac{9(n-1)}{4}=3(n-2)$

$\Rightarrow 3n-3=4n-8$

$\Rightarrow n=5$

Assertion (A) : The coefficient of

$x^{-2}$ in the expansion of

$(x^{2}+\displaystyle \frac{1}{x})^{5}$ is equal to

${ ^{ 5 }C_{ 4 } }$
Reason (R) : The value of r for the above expansion is 3.

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Both A and R are true and R is the correct explanation of A
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Both A and R are true and R is not the correct explanation of A
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A is true, but R is false
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A is false, but R is true

Explanation

For

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

$T_{r+1}=\:^5C_{r}x^{10-3r}$ ...(i)
For the coefficient of

$x^{-2}$

$10-3r=-2$

$3r=12$

$r=4$
Substituting in equation (i)

$T_{5}=\:^5C_{4}$

If the term containing

$x^3$ in

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

${\dfrac{7}{8}}$ when

$x = -2$ and

$n$ is a positive integer, then

$n =$

0%
7
0%
8
0%
9
0%
10

Explanation

The term containing

$x^3$ is

$T_{4}$

$T_{4}=\:^nC_{3} (\dfrac {2}{n})^3$

$=\:^nC_{3}\dfrac{8}{n}$

$=\dfrac{7}{8}$
Therefore

$\dfrac{n(n-1)(n-2)8}{6n^3}=\dfrac{7}{8}$

$=32(n-1)(n-2)=21n^2$

$=11n^2-96n+64$
Hence

$n=8$ and

$n=\dfrac{8}{11}$
But

$n$ has to be a natural number.
Therefore

$n=8$

In the expansion of

$\left[2^{\log_2 {\sqrt{9^{x-1}+7}}}+\displaystyle \frac{1}{2^{\frac{1}{5}\log_2 \displaystyle {(3^{x-1}+1})}}\right]^{7}$ , 6th term is

$84$ . Then

$x =$

0%
$1$
0%
$2$
0%
$1$ or $2$
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$2$ or $4$

Explanation

The given expression

$\displaystyle ={ \left[ \sqrt { { 9 }^{ x-1 }+7 } +\frac { 1 }{ { \left( { 3 }^{ x-1 }+1 \right) }^{ \frac { 1 }{ 5 } } } \right] }^{ 7 }$
Given,

${ T }_{ 6 }=84$

$\Rightarrow\displaystyle {}^{ 7 }{ { C }_{ 5 } }{ \left( \sqrt { { 9 }^{ x-1 }+7 } \right) }^{ 7-5 }{ \left( \frac { 1 }{ { \left( { 3 }^{ x-1 }+1 \right) }^{ \frac { 1 }{ 5 } } } \right) }^{ 5 }=84$

$\displaystyle\Rightarrow {}^{ 7 }{ { C }_{ 5 } }\left( { 9 }^{ x-1 }+7 \right) .\frac { 1 }{ \left( { 3 }^{ x-1 }+1 \right) } =84$

$\Rightarrow { 9 }^{ x-1 }+7=4\left( { 3 }^{ x-1 }+1 \right)$

$\Rightarrow { 3 }^{ 2x }-12.{ 3 }^{ x }+27=0$

$\Rightarrow \left( { 3 }^{ x }-3 \right) \left( { 3 }^{ x }-9 \right) =0$

$\Rightarrow { 3 }^{ x }=3,9$

$\Rightarrow x=1,2.$

$\displaystyle \sum_{{r}=0}^{{n}}({r}-4){C}_{{r}=}$

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$(n-8)2^n$
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$(n-8)2^{n-1}$
0%
$(n-8)2^{n-4}$
0%
$0$

Explanation

In some questions, substituting

$n=a$ positive number in both the question and the answer is the fastest way to achieve the correct option. Although there is always alternate options like writing the general term and splitting it to a format that can be solved but that takes long in a limited time paper. Try to put

$n=2$ . In the end only option B remains.

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

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

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

Explanation

In the following binomial expansion there would total of

$2n-1+1$ terms

$=2n$ number of terms.
Since the total number of terms would be even, there would be two middle terms.

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

$\displaystyle \frac{2n-1+1}{2}+1$ th term.
Hence

$n^{th}$ and

$(n+1)^{th}$ terms.
Coefficient of

$n^{th}$ term

$=T_{(n-1)+1}$

$=\:^{2n-1}C_{n-1}$ ...(i)
Similarly, coefficient of

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

$=T_{(n)+1}$

$=\:^{2n-1}C_{n}$ ...(i)
Adding (i) and (ii), we get

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

$=^{2n}C_{n}$ ...(from identities of binomial coefficients)

$=\displaystyle \frac{2n!}{n!\:n!}$

$=\displaystyle \frac{2n!}{(n!)^{2}}$

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

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

Explanation

In some questions, substituting n=a positive number in both the question and the answer is the fastest way to achieve the correct option. Although there is always alternate options like writing the general term and splitting it to a format that can be solved but that takes long in a limited time paper. Try to put

$n=1$ . In the end only option A remains.

The number of irrational terms in the expansion

${ \left( \sqrt [ 4 ]{ 3 } +\sqrt [ 3 ]{ 7 } \right) }^{ 36 }$ is

0%
$30$
0%
$33$
0%
$31$
0%
$29$

Explanation

Number of rational terms

$=\dfrac{36}{LCM(3,4)}+1$

$=\dfrac{36}{12}+1$

$=4$
Therefore number of irrational terms are

$=$
Total number of terms

$-$ number of rational terms

$=37-4$

$=33$ irrational terms

$\displaystyle \sum_{{r}=0}^{{n}}{r}.{C}_{{r}=}^{2}$

0%
$\dfrac{n}{2}\dfrac{(2n)!}{(n!)^{2}}$
0%
$\dfrac{(2n)!}{(n!)^{2}}$
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$(2n)!$
0%
$\displaystyle \frac{\mathrm{n}(2\mathrm{n})!}{2}$

Explanation

$n=2$ . In the end only option A remains.

The sum of the coefficients of the first

$10$ terms in the expansion of

$(1-x)^{-3}$

0%
$220$
0%
$286$
0%
$120$
0%
$150$

Explanation

For

$(1-x)^{-3}$ , the sum of the first r terms will be

$\:^{n+r-1}C_{n}$
Replacing

$n=3$ and

$r=10$ in the above formula, we get

$\:^{3+10-1}C_{3}$

$=\:^{12}C_{3}$

$=\dfrac{12.11.10}{3!}$

$=\dfrac{12.11.10}{6}$

$=2.11.10$

$=220$

$\displaystyle \sum_{{r}=2}^{{n}}{ ^{ 5r-3 }C_{ r } }=$

0%
$(5{n}+6).2^{{n}-1}-2{n}+2$
0%
$(5{n}+6).2^{{n}-1}-2{n}+3$
0%
$(5{n}-6)2^{{n}-1}-2{n}+2$
0%
$(5{n}-6)2^{{n}-1}-2{n}+3$

Explanation

The given question can be re-framed as

$5\sum_{r=2} ^{n}rC_{r}-3\sum_{r=2} ^{n}C_{r}$

$5[n2^{n-1}-n]-3[2^n-n-1]$

$=2^{n-1}(5n-6)-2n+3$

$S_{1}=^{m}{c}_{1}+(m+1)_{C_{2}+}(m+2)_{C_{3}+\ldots.+}(m+n-1)_{C_{n}}$

$S_{2}=^{n}{c}_{1}+(n+1)_{C_{2}+}(n+2)_{C_{3}+\ldots.+}(m+n-1)_{C_{n}}$

0%
${S}_{1}+{S}_{2}=0$
0%
${S}_{1}-{S}_{2}=0$
0%
$S_1+S_2=2^n$
0%
${S}_{1}+{S}_{2}=2^{{n}}-1$

Explanation

$S_1$ =

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

$^{m+1}{C}_{2}$ +

$^{m+2}{C}_{3}$ +........+

$^{m+n-1}{C}_{n}$

Adding 1 both sides

$S_1$ +1=

$^{m}{C}_{0}$ +

$^{m}{C}_{1}$ ........... using (

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

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

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

$S_1$ +1=

$^{m+1}{C}_{1}$ +

$^{m+1}{C}_{2}$ .........

Similarly, this can seen that

$S_1$ +1=

$^{m+n-1}{C}_{n-1}$ +

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

$^{m+n}{C}_{n}$ eqn 1

Similarly,it can be seen that

$S_2$ +1=

$^{m+n}{C}_{n}$ eqn 2

(eqn 1)-(eqn 2)

(

$S_1$ +1)-(

$S_2$ +1)=

$^{m+n}{C}_{n}$ -

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

$S_1$ -

$S_2$ =0

$\mathrm{C}_{0}+2\mathrm{C}_{1}+4\mathrm{C}_{2}+\ldots.+2^{\mathrm{n}}\mathrm{C}_{\mathrm{n}}=243$ , then n =

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

The sum

$\displaystyle \sum_{\lrcorner 0<\leq}\sum_{{j}\leq 10}{C}_{{j}}({C}_{{i}})=$

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

Explanation

The above question can be written as

$\:^{10}C_{1}(\:^{1}C_{0}+\:^{1}C_{1})+\:^{10}C_{2}(\:^{2}C_{0}\:^{2}C_{1}\:^{2}C_{2})+...$

$=\:^{10}C_{1}(2)+\:^{10}C_{2}(2^2)+\:^{10}C_{2}(2^2)+\:^{10}C_{2}(2^2)...+\:^{10}C_{10}(2^{10})$
Since

$\:^{10}C_{n}=\:^{10}C_{10-n}$
Therefore

$=\:^{10}C_{1}(2^1+2^9)+\:^{10}C_{2}(2^2+2^8)+\:^{10}C_{3}(2^3+2^7)+\:^{10}C_{4}(2^4+2^6)+\:^{10}C_{5}(2^5)+\:^{10}C_{10}(2^{10})$

$=10(2^1+2^9)+45(2^2+2^8)+120(2^3+2^7)+210(2^4+2^6)+252(2^5)+2^{10}$

$=59048$

$=59049-1$

$=3^{10}-1$

The number of non zero terms in the expansion of

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

0%
$9$
0%
$0$
0%
$5$
0%
$10$

Explanation

$(1+3\sqrt{2}x)\:^{9}+(1-3\sqrt{2}x)\:^{9}$
Applying binomial theorem for expansion of the following

$=1+(3\sqrt{2}x)\:(^{9}C_{1})+(3\sqrt{2}x)^{2}\:(^{9}C_{2})...(\:^{9}C_{9}(3\sqrt{2}x)^{9})$

$+1-(3\sqrt{2}x)\:(^{9}C_{1})+(3\sqrt{2}x)^{2}\:(^{9}C_{2})...-(\:^{9}C_{9}(3\sqrt{2}x)^{9})$

$=2[1+(3\sqrt{2}x)\:^2\:(^{9}C_{2})+(3\sqrt{2}x)\:^4\:(^{9}C_{4})...(3\sqrt{2}x)\:^8(\:^{9}C_{8})]$
The powers of

$3\sqrt{2}x$ are in A.P

$0,2,4,6,8$
Hence, including the term

$1\:or\: x^0$ there are

$\dfrac{9+1}{2}$ terms.
Hence, there are

$5$ terms,

$x + y = 1$ , then

$\displaystyle \sum_{r=0}^{n}r^{n}C_{r}x^{r}.y^{n-r}=$

0%
$1$
0%
$n$
0%
$nx$
0%
$ny$

Explanation

$\displaystyle \sum_{r=0} ^{n}r\:^{n}C_{r}x^ry^{n-r}$

$=\displaystyle \sum_{r=0} ^{n}r\frac{n!}{(n-r)!r!}x^ry^{n-r}$

$=nx(y+x)^{n-1}$

$=nx(1)^{n-1}$ ... given in the above question

$=nx$
Hence, the answer is Option C

$x + y = 1$ then

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

$^{n}C_{r}x^{r}y^{n-r}$

0%
$nxy$
0%
$nx(x + yn)$
0%
$nx(nx + y)$
0%
$nx$

Explanation

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

$^{n}C_{r}x^{r}y^{n-r}$

$\displaystyle=\sum_{r=0} ^{n}[r(r-1)+r]$

$^{n}C_{r}x^ry^{n-r}$

$\displaystyle =\sum_{r=0} ^{n}(r(r-1))\cdot\ ^{n}C_{r}x^ry^{n-r}+\sum_{r=0}^{n}r\cdot\ ^{n}C_{r}x^ry^{n-r}$

$\displaystyle\sum_{r=0} ^{n}(r(r-1))\cdot\ ^{n}C_{r}x^ry^{n-r} = x^2 \dfrac{d^2}{dx^2}(\sum_{r=0}^{n}$

$^{n}C_{r}x^ry^{n-r}) = x^2\dfrac{d^2}{dx^2}(x+y)^n = n(n - 1)x^2$

$\displaystyle\sum_{r=0}^{n}r\cdot\ ^{n}C_{r}x^ry^{n-r} = x \dfrac{d}{dx}(\sum_{r=0}^{n}$

$^{n}C_{r}x^ry^{n-r}) = x\dfrac{d}{dx}(x+y)^n = nx$

$\therefore\displaystyle \sum_{r=0} ^{n}(r(r-1))\cdot\ ^{n}C_{r}x^ry^{n-r}+\sum_{r=0}^{n}r\cdot\ ^{n}C_{r}x^ry^{n-r}$

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

$=nx[x(n-1)+1]$

$=nx[nx-x+(x+y)]$ ...it is given in the question that

$x+y=1$

$=nx[nx+y]$
Hence, the answer is Option C

The number of non zero terms in

$(x+\mathrm{a})^{75}+(x-\mathrm{a})^{75}$

0%
$38$
0%
$76$
0%
$34$
0%
$32$

Explanation

$(x+a)\:^{75}+(x-a)\:^{75}$
Applying binomial theorem for expansion of the following

$(x+a)\:^{75}+(x-a)\:^{75}$

$=x^{75}+x^{74}\:(^{75}C_{1}a)+x^{73}\:(^{75}C_{2}a^2)...(\:^{75}C_{75}a^{75})$

$+x^{75}-x^{74}\:(^{75}C_{1}a)+x^{73}\:(^{75}C_{2}a^2)...(\:^{75}C_{75}a^{75})$

$=2[x^{75}+x^{73}\:(^{75}C_{2}a^2)+x^{71}\:(^{75}C_{4}a^4)...(\:^{75}C_{74}a^{74}x)]$
The powers of

$a$ are in A.P

$a^0,a^2,a^4,a^6...a^{74}$
Therefore

$a_{n}=a_{1}+(n-1)d$

$74=0+(n-1)2$

$37=n-1$

$38=n$
Hence, including the term

$a^0\:or\:x^{75}$ there are

$38$ terms.
Hence, answer is

$38.$

If the fourth term in the expansion of

$\left(\sqrt{x^\dfrac{1}{\log x+1}}+x^\dfrac{1}{12}\right)^{6}$ is equal to

$200$ and

$x>1$ , then

$x$ is

0%
$10$
0%
$10^{-4}$
0%
$1$
0%
$-4$

Explanation

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$
Applying to the above question, we get
$T_{3+1}=\:^6C_{3}x^{\dfrac {3}{2(\log x+1)}}x^{\dfrac {3}{12}}$
$=200$
$20x^{\dfrac {3}{2(\log x+1)}}x^{\dfrac {3}{12}}=200$
$x^{\dfrac {3}{2(\log x+1)}+\dfrac {1}{4}}=10$
Taking $\log$ to the base $10$ ( $\log_{10}$ ) on both sides, we get
$\dfrac{3}{2(\log x+1)}+\dfrac{1}{4}=1$
$\dfrac{3}{2(\log x+1)}=\dfrac{3}{4}$
$4=2(\log_{10}x+1)$
$2=\log_{10}x+1$
$1=\log_{10}x$
Taking anti-logarithm, we get
$x=10$

If in the expansion of

$(\displaystyle \frac{1}{x}+x\tan x)^{5}$ , the ratio of

$4^{th}$ term to the

$2^{nd}$ term is

$\displaystyle \frac{2}{27}\pi^{4}$ , then the value of

${x}$ can be

0%
$\displaystyle \frac{-\pi}{6}$
0%
$\displaystyle \frac{-\pi}{3}$
0%
$\displaystyle \frac{\pi}{3}$
0%
$\displaystyle \frac{\pi}{12}$

Explanation

From the above given expression by applying binomial theorem, we get.

$T_{4}=\:^5C_{3}x\tan^3{x}$

$=10x\tan^3{x}$

$T_{2}=\:^5C_{1}x^{-3}\tan{x}$

$=5x^{-3}\tan{x}$
Therefore

$\displaystyle \frac{T_{4}}{T_{2}}$

$=\displaystyle \frac{10x\tan^3{x}}{5x^{-3}\tan{x}}$

$=2x^4\tan^2x$

$=\displaystyle \frac{2\pi^{4}}{27}$

$x^4\tan^2x=\displaystyle \frac{\pi^{4}}{27}$
Taking root on both sides, we get

$x^2\tan x=\displaystyle \frac{\pi^2}{3\sqrt3}$

$=\displaystyle \frac{\sqrt{3}\pi^2}{9}$
Therefore

$x^2=\displaystyle \frac{\pi^2}{9}$ ...(i) and

$\tan x=\sqrt{3}$ ...(ii)
Both Eq(i) and Eq (ii) give

$x=\displaystyle \frac{\pi}{3}$

In the expansion of

$(1+x)^{n}.(1+y)^{n}.(1+\mathrm{z})^{n}$ the sum of the coefficients of the terms of degree

$r$ is

0%
$(^{n}C_{r})^{3}$
0%
$^{3n}C_{r}$
0%
$3\times nC_{r}$
0%
$nC_{3r}$

Explanation

The given expression contains

$3n$ factors
Using combination to choose

$r$ brackets out of

$3n$ brackets for a term of degree

$r$ , we get

$\:^{3n}C_r$
Hence, answer is option B

The number of terms in the expansion of

$\left [ (a+4b)^{3}(a-4b)^{3} \right ]^{2}$ are

0%
$6$
0%
$7$
0%
$8$
0%
$32$

Explanation

$[(a+4b)\:^3(a-4b)\:^3]\:^2$

$=[(a+4b)(a-4b)]\:^6$

$=[a^2-16b^2]\:^6$
Hence total number of terms is

$n+1$
Here

$n=6$
Therefore, total number of terms is

$7.$

The number of terms in the expansion of

$\left [ (a+4b)^{3}+(a-4b)^{3} \right ]^{2}$ are

0%
$6$
0%
$8$
0%
$7$
0%
$3$

Explanation

$(x^3+y^3)$

$=(x+y)\:^3-3xy(x+y)$
Applying the above formula on the given question we get

$[(a+4b)\:^3+(a-4b)\:^3]\:^2$

$=[8a^3-6a(a^2-16b^2)]\:^2$

$=[96ab^2+2a^3]\:^2$

$=4a^2[48b^2+a^2]\:^2$

$=4[2304a^2b^4 +96a^4b^2 +a^6]$
Hence, the total number of terms is

$3$ .

The sum of the coefficients of the middle terms of

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

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

Explanation

Consider:

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

Since

$2n-1$ is odd, the middle terms are

$\left(\dfrac{2n-1+1}{2}\right)^{th}$ and

$\left(\dfrac{2n-1+1}{2}+1\right)^{th}$ terms
Now, consider the following

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(i)
Where

$T$ represents the term
Here the middle terms are the

$n^{th}$ term and

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

$T_{n+1}=\ ^{2n-1}C_{n}1^{n-1}x^{n}$

$=\:^{2n-1}C_{n}x^{n}$
Hence, the coefficient is

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

$T_{n}=\ ^{2n-1}C_{n-1}1^{n}x^{n-1}$

$=\:^{2n-1}C_{n-1}x^{n-1}$
Hence,

$\:^{2n-1}C_{n-1}$ is the coefficient ...(2)

Therefore, sum of the coefficients of the middle terms is

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

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

Hence, option D.

The coefficient of

$x^4$ in

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

0%
$\displaystyle \frac{45}{64}$
0%
$\displaystyle \frac{243}{128}$
0%
$\displaystyle \frac{405}{256}$
0%
$\displaystyle \frac{810}{512}$

Explanation

$\displaystyle \frac{10 \times 1- 4}{1+2} = \frac{10-4}{3} = \frac{6}{3}=2$
Coefficient of

$x^4$ is

$^{10}C_2 \left ( \frac{1}{2} \right )^{10-2}(-3)^2$

$\displaystyle \frac{10 \times 9}{2} \cdot \frac{1}{2^8} \cdot 3^2 = \frac{10 \times 81}{2 \times 2^8}$

$\displaystyle \frac{5 \times 81}{256} = \frac{405}{256}$

If the coefficients of

$r^{th}$ term and

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

$(1+x)^{20}$ are in the ration 1 : 2, then

$r=$

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

Explanation

The ratio of the coefficient of rth term and (r+1)th term 1 : 2

$\Rightarrow ^{20}C_{r-1} : ^{20}C_r = 1: 2$

$\Rightarrow \displaystyle \frac{r}{21-r} = \frac{1}{2} \Rightarrow 2r = 21-r$

$\Rightarrow r =7$

The coefficient of the

$8$ th term in the expansion of

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

0%
$120$
0%
$7$
0%
$^{10}C_8$
0%
$210$

Explanation

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

$^{10}C_0+\,^{10}C_1x+\,^{10}C_2x^2+........+\,^{10}C_7x^7+\,^{10}C_8x^8+\,^{10}C_9x^9+\,^{10}C_{10}x^{10}$

So here, first term is

$^{10}C_0$ then

$8^{th}$ term will be

$10C_7x^7.$

$\Rightarrow$ Coefficient of the

$8^{th}$ term

$=\,^{10}C_7$

$=\dfrac{10!}{7!3!}$

$=\dfrac{10\times 9\times 8\times 7!}{7!\times 3\times 2\times 1}$

$=120$

$T_r$ denotes the rth term in the expansion of

$\displaystyle \left ( x+\frac{1}{y} \right)^{23}$ then

0%
$T_{12}=T_{13}$
0%
$x^2T_{13}=T_{12}$
0%
$T_{12} = xy T_{13}$
0%
$T_{12} + T_{12} = 25$

Explanation

$T_{12} = ^{23} C_{11}x^{12} \displaystyle \left ( \frac{1}{y} \right)^{11} = ^{23}C_{11} \frac{x^{12}}{y^{11}}$ ,

$T_{13}=^{23}C_{12} x^{11} \displaystyle \left ( \frac{1}{y} \right)^{12} = ^{23}C_{11}\frac{x^{11}}{y^{12}}$ ,

$T_{12} = xy \displaystyle \left ( ^{23}C_{11} \frac{x^{11}}{y^{12}} \right ) = xy T_{13}$

The coefficient of

$x^3$ in

$\displaystyle \left ( \sqrt{x^5}+ \frac{3}{\sqrt{x^3}} \right )^5$ is

0%
0
0%
120
0%
420
0%
540

Explanation

$\displaystyle r=\frac{6 \times \frac{5}{2}-3}{\frac{5}{2}+\frac{3}{2}} = \frac{15-3}{4} =3$

$\therefore$ Coefficient of

$x^3$ is

$^6C_3 3^3$

$\displaystyle =\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \cdot 27$

$=5 \times 4 \times 27=540$

The coefficient of the middle term in the expansion of

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

0%
$^{2n}C_{n}$
0%
$\displaystyle \frac{1.3.5\ldots..(2n-1)}{n!}2^{n}$
0%
$2.6\ldots(4n-2)$
0%
$2.4\ldots\ldots\ldots 2n$

Explanation

Middle term

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

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

$=(n+1)th\:term$

Now consider the following

$T_{r+1}=\:^nC_{r}a^{n-r}b^r$ ...(i)

Where

$T$ represents the term.

Here the middle term is the

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

Hence

$r+1=n+1$

$r=n$

substituting in (i), we get

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

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

Hence the coefficient of middle term is

$\:^{2n}C_{n}$ which can also be written as

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

Assertion (A) : Number of the disimilar terms in the sum of expansion $(x+a)^{102}+(x-a)^{102}$ is $206$

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

0%
Both A and R are individually true and R is the correct explanation of A.
0%
Both A and R are individually true and R is not correct explanation of A.
0%
A is true but R is false
0%
A is false but R is true

Explanation

$(x+a)^{102}+(x-a)^{102}$
Since a is constant let

$a=1$
Therefore the above question can be re-written as

$(1+x)^{102}+(1-x)^{102}$

$=[1+\:^{102}C_{1}x^1+\:^{102}C_{2}x^2...+\:^{102}C_{102}x^{102}]+[1-\:^{102}C_{1}x^1+\:^{102}C_{2}x^2-\:^{102}C_{3}x^3...+\:^{102}C_{102}x^{102}]$

$=2[1+\:^{102}C_{2}x^2+\:^{102}C_{2}x^4...+\:^{102}C_{102}x^{102}]$ ...(i)
Hence number of dissimilar terms in Eq(i) will be

$\dfrac{102}{2}+1$

$=52$ terms.
Hence A is false.
Reason is true.
Hence option D is the correct option.

In the binomial expansion of

$(a-b)^n, n \geq 5$ , the sum of

$5^{th}$ and

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

$\dfrac ab$ equals

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

Explanation

Given

$T_5+T_6=0$

$\Rightarrow ^nC_4 a^{n-4} \cdot b^4 - ^nC_5 a^{n-5}b^5=0$

$\Rightarrow ^nC_4 a^{n-4}b^4=^nC_5 a^{n-5}b^5$

$\Rightarrow \displaystyle \frac{a^{n-4}b^4}{a^{n-5}b^5}=\frac{^nC_5}{^nC_4}$

$\Rightarrow \displaystyle \frac{a}{b} = \frac{n-5+1}{5} \Rightarrow \frac{a}{b} = \frac{n-4}{5}$

The total number of rational terms in the expansion of

$\left(7^{\frac 13} + 11^{\frac 19}\right)^{6561}$ is

0%
$731$
0%
$729$
0%
$728$
0%
$730$
0%
$732$

Explanation

Total number of rational terms will be

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

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

$=729+1$

$=730$

If the coefficient of

$x^7$ in

$\displaystyle \left [ ax^2 + \left ( \dfrac{1}{bx} \right ) \right ]^{11}$ equals the coefficient of

$x^{-7}$ in

$\displaystyle \left [ ax - \left ( \dfrac{1}{bx^2} \right ) \right ]^{11}$ , then

$a$ and

$b$ satisfy the relation

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

Explanation

Coefficient of

$x^7$ in

$\displaystyle \left ( ax^2 + \frac{1}{bx} \right )^{11}$ is

$^{11}C_5 a^6 \dfrac{1}{b^5}$
Coefficient of

$x^{-7}$ in

$\displaystyle \left ( ax-\frac{1}{bx^2} \right )^1$ is

$^{11}C_6 a^5 \displaystyle \frac{1}{b^6}$

$^{11}C_6 a^5 \displaystyle \frac{1}{b^6}={}^{11}C_5 a^6 \displaystyle \frac{1}{b^5}$

$\Rightarrow ab = \displaystyle \frac{^{11}C_6}{^{11}C_5} = \frac{6}{6}=1$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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