Binomial Theorem — Page 10 | JEE Mathematics

Q91

Let the coefficients of three consecutive terms

T_r

,

T_{r+1}

and

T_{r+2}

in the binomial expansion of

(a + b)^{12}

be in a G.P. and let

p

be the number of all possible values of

r

. Let

q

be the sum of all rational terms in the binomial expansion of

(\sqrt[4]{3}+\sqrt[3]{4})^{12}

. Then

p + q

is equal to:

A 295

B 283

C 299

D 287

Correct Answer

Option B

Solution

\begin{aligned} & (\mathrm{a}+\mathrm{b})^{\frac{1}{2}} \\ & \mathrm{~T}_{\mathrm{r}}, \mathrm{~T}_{\mathrm{r}+1}, \mathrm{~T}_{\mathrm{r}+2} \rightarrow \mathrm{GP} \\ & \text{ So, } \frac{\mathrm{T}_{\mathrm{r}+1}}{\mathrm{~T}_{\mathrm{r}}}=\frac{\mathrm{T}_{\mathrm{r}+2}}{\mathrm{~T}_{\mathrm{r}+1}} \\ & \frac{{ }^{12} \mathrm{C}_{\mathrm{r}}}{{ }^{12} \mathrm{C}_{\mathrm{r}-1}}=\frac{{ }^{12} \mathrm{C}_{\mathrm{r}+1}}{{ }^{12} \mathrm{C}_{\mathrm{r}}} \\ & \frac{12-\mathrm{r}+1}{\mathrm{r}}=\frac{12-(\mathrm{r}+1)+1}{\mathrm{r}+1} \\ & (13-\mathrm{r})(\mathrm{r}+1)=(12-\mathrm{r})(\mathrm{r}) \\ & -\mathrm{r}+12 \mathrm{r}+13=12 \mathrm{r}-\mathrm{r}^2 \\ & 13=0 \end{aligned}

No value of r possible So $\mathrm{P}=0$

\left(3^{\frac{1}{4}}+4^{\frac{1}{3}}\right)^{12}=\sum{ }^{12} C_r\left(3^{\frac{1}{4}}\right)^{12-\mathrm{r}}\left(4^{\frac{1}{3}}\right)^{\mathrm{r}}

Exponent of $\left(3^{\dfrac{1}{4}}\right)$ exponent of $\left(4^{\dfrac{1}{3}}\right)$ term

\begin{array}{ccc} 12 & 0 & 27 \\ 0 & 12 & 256 \\ \mathrm{q}=27+256=283 & & \\ \mathrm{p}+\mathrm{q}=0+283=283 & & \end{array}

Q92

If in the expansion of

(1+x)^{\mathrm{p}}(1-x)^{\mathrm{q}}

, the coefficients of

x

and

x^2

are 1 and -2 , respectively, then

\mathrm{p}^2+\mathrm{q}^2

is equal to :

A 8

B 20

C 13

D 18

Correct Answer

Option C

Solution

\begin{aligned} & (1+\mathrm{x})^{\mathrm{p}}(1-\mathrm{x})^{\mathrm{q}}=\left({ }^{\mathrm{p}} \mathrm{C}_0+{ }^{\mathrm{p}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{p}} \mathrm{C}_2 \mathrm{x}^2+\ldots\right)\left({ }^q \mathrm{C}_0-{ }^q \mathrm{C}_1 \mathrm{x}+{ }^q \mathrm{C}_2 \mathrm{x}^2+\ldots\right) \\ & \text{ coff of } \mathrm{x} \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^{\mathrm{q}} \mathrm{C}_1-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_0=1 \\ & \mathrm{p}-\mathrm{q}=1 \\ & \text{ coff of } \mathrm{x}^2 \equiv{ }^{\mathrm{p}} \mathrm{C}_0{ }^q \mathrm{C}_2-{ }^{\mathrm{p}} \mathrm{C}_1{ }^q \mathrm{C}_1+{ }^{\mathrm{p}} \mathrm{C}_2{ }^{\mathrm{q}} \mathrm{C}_0=-2 \\ & \frac{\mathrm{q}(\mathrm{q}-1)}{2}-\mathrm{pq}+\frac{\mathrm{p}(\mathrm{p}-1)}{2}=-2 \\ & \mathrm{q}^2-\mathrm{q}-2 \mathrm{pq}+\mathrm{p}^2-\mathrm{p}=-4 \\ & (\mathrm{p}-\mathrm{q})^2-(\mathrm{p}+\mathrm{q})=-4 \\ & \mathrm{p}+\mathrm{q}=5 \\ & \mathrm{p}=3 \\ & \mathrm{q}=2 \\ & \text{ so } \mathrm{p}^2+\mathrm{q}^2=13 \end{aligned}

Q93

If

\sum\limits_{r=1}^9\left(\dfrac{r+3}{2^r}\right) \cdot{ }^9 C_r=\alpha\left(\dfrac{3}{2}\right)^9-\beta, \alpha, \beta \in \mathbb{N}

, then

(\alpha+\beta)^2

is equal to

A 27

B 81

C 18

D 9

Correct Answer

Option B

Solution

\begin{aligned} & \sum_{r=1}^9\left(\frac{r+3}{2^r}\right) \cdot{ }^9 C_r=\sum_{r=1}^9 \frac{r}{2^r} \cdot \frac{9}{r} \cdot{ }^8 C_{r-1}+\sum_{r=1}^9 3 \cdot{ }^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2} \sum_{r=1}^9{ }^8 C_{r-1}\left(\frac{1}{2}\right)^{r-1}+3 \sum_{r=1}^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2} \sum_{r=0}^8{ }^8 C_{r-1}\left(\frac{1}{2}\right)^r+3 \sum_{r=1}^9{ }^9 C_r\left(\frac{1}{2}\right)^r \\ & =\frac{9}{2}\left(1+\frac{1}{2}\right)^8+3\left[\left(1+\frac{1}{2}\right)^9-{ }^9 C_0\left(\frac{1}{2}\right)^0\right] \\ & =\frac{9}{2} \cdot \frac{3^8}{2^8}+3\left[\frac{3^9}{2^9}-1\right] \\ & =\frac{3^{10}}{2^9}+\frac{3^{10}}{2^9}-3=4 \cdot \frac{3^{10}}{2^{10}}-3 \\ & =4\left(\frac{3}{2}\right)^{10}-3 \\ & =6\left(\frac{3}{2}\right)^9-3 \\ & \alpha=6, \beta=3 \Rightarrow(\alpha+\beta)^2=81 \end{aligned}

Q94

For some

\mathrm{n} \neq 10

, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of

(1+\mathrm{x})^{\mathrm{n}+4}

be in A.P. Then the largest coefficient in the expansion of

(1+\mathrm{x})^{\mathrm{n}+4}

is:

A 10

B 35

C 70

D 20

Correct Answer

Option B

Solution

\begin{aligned} & (1+\mathrm{x})^{\mathrm{n+4}} \\ & { }^{\mathrm{n}+4} \mathrm{C}_4{ }^{\mathrm{n+4}} \mathrm{C}_5{ }^{\mathrm{n}+4} \mathrm{C}_6, \rightarrow \text{ A.P. } \\ & \Rightarrow 2 \times{ }^{\mathrm{n}+4} \mathrm{C}_5={ }^{\mathrm{n+4}} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_6 \\ & \Rightarrow 4 \times{ }^{\mathrm{n+4}} \mathrm{C}_5=\left({ }^{n+4} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_5\right)+\left({ }^{\mathrm{n+4}} \mathrm{C}_5+{ }^{\mathrm{n}+4} \mathrm{C}_6\right) \\ & \Rightarrow 4 \times{ }^{n+4} \mathrm{C}_5={ }^{\mathrm{n}+5} \mathrm{C}_5+{ }^{\mathrm{n}+5} \mathrm{C}_6 \\ & \Rightarrow 4 \times \frac{(\mathrm{n}+4)!}{5!.(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\mathrm{n}!} \\ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \\ & \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \\ & \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \\ & \Rightarrow \mathrm{n}=3,10(\text{ rejected }) \\ & \because \mathrm{n} \neq 10 \end{aligned}

$\therefore$ Largest binomial coefficient in expansion of

\begin{aligned} & (1+x)^7 \\ & (\because \mathrm{n}+4=7) \end{aligned}

is coeff. of middle term

\Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35

N.T.A. Ans Option (2)

Q95

Suppose

A

and

B

are the coefficients of

30^{\text{th }}

and

12^{\text{th }}

terms respectively in the binomial expansion of

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

. If

2 \mathrm{~A}=5 \mathrm{~B}

, then n is equal to:

A 20

B 19

C 22

D 21

Correct Answer

Option D

Solution

\begin{aligned} & A={ }^{2 n-1} C_{29} \quad B={ }^{2 n-1} C_{11} \\ & 2{ }^{2 n-1} C_{29}=5{ }^{2 n-1} C_{11} \\ & 2 \frac{(2 n-1)!}{29!(2 n-30)!}=5 \frac{(2 n-1)!}{(2 n-12)!11!} \\ & \frac{1}{29 \ldots 12 \cdot 5}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 2} \\ & \frac{1}{30 \cdot 29 \ldots 12}=\frac{1}{(2 n-12)(2 n-13) \ldots(2 n-29) 12} \\ & 2 n-12=30 \\ & n=21 \end{aligned}

Q96

The number of integral terms in the expansion of

\left( {5^\dfrac{1}{2}} + 7^\dfrac{1}{8} \right)^{1016}

is:

A 127

B 128

C 130

D 129

Correct Answer

Option B

Solution

\begin{aligned} &\begin{aligned} & \mathrm{T}_{\mathrm{r}}={ }^{1016} \mathrm{C}_{\mathrm{r}}(5)^{\frac{1016-\mathrm{r}}{2}} 7^{\frac{\mathrm{r}}{8}} \\ & \Rightarrow \mathrm{r}=0,8,16,24, \ldots ., 1016 \\ & 1016=0+(\mathrm{n}-1) 8 \\ & \Rightarrow \mathrm{n}-1=\frac{1016}{8}=127 \end{aligned}\\ &\text{ So, } \mathrm{n}=128 \text{. } \end{aligned}

Q97

The largest

\mathrm{n} \in \mathbf{N}

such that

3^{\mathrm{n}}

divides 50 ! is :

A 22

B 20

C 21

D 23

Correct Answer

Option A

Solution

\begin{aligned} & V_3(50!)=\left[\frac{50}{3}\right]+\left[\frac{50}{9}\right]+\left[\frac{50}{27}\right]+\left[\frac{50}{81}\right]+\ldots+\left[\frac{50}{3^n}\right] \\ & =16+5+1+0+0+0 \ldots=22 \end{aligned}

Q98

The term independent of

x

in the expansion of

\left(\dfrac{(x+1)}{\left(x^{2 / 3}+1-x^{1 / 3}\right)}-\dfrac{(x-1)}{\left(x-x^{1 / 2}\right)}\right)^{10}, x>1

, is :

A 240

B 120

C 150

D 210

Correct Answer

Option D

Solution

{\left( {{{x + 1} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{x - 1} \over {x - {x^{1/2}}}}} \right)^{10}}

=

{\left( {{{{{\left( {{x^{1/3}}} \right)}^3} + {{\left( {{1^{1/3}}} \right)}^3}} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{{{\left( {\sqrt x } \right)}^2} - {{\left( 1 \right)}^2}} \over {x - {x^{1/2}}}}} \right)^{10}}

=

{\left( {{{\left( {{x^{1/3}} + 1} \right)\left( {{x^{2/3}} - {x^{1/3}} + 1} \right)} \over {{x^{2/3}} - {x^{1/3}} + 1}} - {{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)} \over {\sqrt x \left( {\sqrt x - 1} \right)}}} \right)^{10}}

=

{\left( {\left( {{x^{1/3}} + 1} \right) - {{\left( {\sqrt x + 1} \right)} \over {\sqrt x }}} \right)^{10}}

=

{\left( {\left( {{x^{1/3}} + 1} \right) - \left( {1 + {1 \over {\sqrt x }}} \right)} \right)^{10}}

=

{\left( {{x^{1/3}} - {1 \over {{x^{1/2}}}}} \right)^{10}}

[Note: For

{\left( {{x^\alpha } \pm {1 \over {{x^\beta }}}} \right)^n}

the

\left( {r + 1} \right)

th term with power m of x is

r = {{n\alpha - m} \over {\alpha + \beta }}

] Here

\alpha = {1 \over 3}

,

\beta = {1 \over 2}

and m = 0 then

r = {{10 \times {1 \over 3} - 0} \over {{1 \over 3} + {1 \over 2}}}

=

{{10} \over 3} \times {6 \over 5}

= 4 $\therefore$ T5 is the term independent of x. $\therefore$ T5 =

{}^{10}{C_4}

= 210

Q99

The remainder when

\left((64)^{(64)}\right)^{(64)}

is divided by 7 is equal to

A 4

B 6

C 3

D 1

Correct Answer

Option D

Solution

\begin{aligned} &\begin{aligned} & 64^{64} \Rightarrow(63+1)^{64}=63 \lambda+1 \\ & 64^{664^{64}} \Rightarrow(63+1)^{64^{64}} \\ & 63 \lambda_1+1 \end{aligned}\\ &\text{ Required remainder when divided } 7 \text{ is } 1 \text{. } \end{aligned}

Q100

If\,\sum\limits_{r = 0}^{10} {({{{{10}^{r + 1}} - 1} \over {{{10}^r}}}).{}^{11}{C_{r + 1}} = {{{}_\alpha 11 - {{11}^{11}}} \over {{{10}^{10}}}},\,then\,\,\alpha \,\,is\,\,equal\,\,to:}

A 11

B 20

C 24

D 15

Correct Answer

Option B

Solution

$\begin{aligned} & \sum_{r=0}^{10}\left(\dfrac{10^{r-1}-1}{10^r}\right){ }^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =\sum_{\mathrm{r}=0}^{10}\left(10-\dfrac{1}{10^r}\right){ }^{11} \mathrm{C}_{\mathrm{r}+1} \\ & =10 \sum_{\mathrm{r}=0}{ }^{11} \mathrm{C}_{\mathrm{r}+1}-10 \sum\left({ }^{11} \mathrm{C}_{\mathrm{r}+1}\left(\dfrac{1}{10}\right)^{r+1}\right) \\ & =10\left[{ }^{11} \mathrm{C}_1+{ }^{11} \mathrm{C}_2+\ldots . .+{ }^{11} \mathrm{C}_{11}\right] \\ & -10\left[{ }^{11} \mathrm{C}_1\left(\dfrac{1}{10}\right)^1+{ }^{11} \mathrm{C}_2\left(\dfrac{1}{10}\right)^2+\ldots . .+{ }^{11} \mathrm{C}_{11}\left(\dfrac{1}{10}\right)^{11}\right] \\ & =10\left[2^{11}-1\right]-10\left[\left(1+\dfrac{1}{10}\right)^{11}-1\right] \\ & =10(2)^{11}-10-\dfrac{11^{11}}{10^{10}}+10 \\ & =\dfrac{(20)^{11}-11^{11}}{10^{10}} \\ & \therefore \alpha=20\end{aligned}$