Binomial Theorem — Page 7 | JEE Mathematics

Q61

Let K be the sum of the coefficients of the odd powers of

x

in the expansion of

(1+x)^{99}

. Let

a

be the middle term in the expansion of

{\left( {2 + {1 \over {\sqrt 2 }}} \right)^{200}}

. If

{{{}^{200}{C_{99}}K} \over a} = {{{2^l}m} \over n}

, where m and n are odd numbers, then the ordered pair

(l,\mathrm{n})

is equal to

A (50, 101)

B (50, 51)

C (51, 101)

D (51, 99)

Correct Answer

Option A

Solution

K = {2^{98}}

a = {}^{200}{C_{100}}\,{2^{50}}

$\therefore$

{{{}^{200}{C_{99}}\,.\,{2^{98}}} \over {{}^{200}{C_{100}}\,.\,{2^{50}}}} = {{{2^l}m} \over n}

\Rightarrow {{100} \over {101}}\,.\,{2^{48}} = {{{2^l}m} \over n}

\Rightarrow {{25} \over {101}}\,.\,{2^{50}} = {{{2^l}m} \over n}

$\therefore$

l = 50,m = 25,n = 101

Q62

If

a_r

is the coefficient of

x^{10-r}

in the Binomial expansion of

(1 + x)^{10}

, then

\sum\limits_{r = 1}^{10} {{r^3}{{\left( {{{{a_r}} \over {{a_{r - 1}}}}} \right)}^2}}

is equal to

A 3025

B 4895

C 5445

D 1210

Correct Answer

Option D

Solution

\begin{aligned} & \mathrm{a}_{\mathrm{r}}={ }^{10} \mathrm{C}_{10-\mathrm{r}}={ }^{10} \mathrm{C}_{\mathrm{r}} \\\\ & \Rightarrow \sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{{ }^{10} \mathrm{C}_{\mathrm{r}}}{{ }^{10} \mathrm{C}_{\mathrm{r}-1}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}^3\left(\frac{11-\mathrm{r}}{\mathrm{r}}\right)^2=\sum_{\mathrm{r}=1}^{10} \mathrm{r}(11-\mathrm{r})^2 \\\\ & =\sum_{\mathrm{r}=1}^{10}\left(121 \mathrm{r}+\mathrm{r}^3-22 \mathrm{r}^2\right)=1210 \end{aligned}

Q63

If

{({}^{30}{C_1})^2} + 2{({}^{30}{C_2})^2} + 3{({}^{30}{C_3})^2}\, + \,...\, + \,30{({}^{30}{C_{30}})^2} = {{\alpha 60!} \over {{{(30!)}^2}}}

then

\alpha

is equal to :

A 30

B 10

C 15

D 60

Correct Answer

Option C

Solution

\begin{aligned} & \mathrm{S}=0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+1 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+2 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots \ldots+30 \cdot\left({ }^{30} \mathrm{C}_{30}\right)^2 \\\\ & \mathrm{S}=30 \cdot\left({ }^{30} \mathrm{C}_0\right)^2+29 \cdot\left({ }^{30} \mathrm{C}_1\right)^2+28 \cdot\left({ }^{30} \mathrm{C}_2\right)^2+\ldots . \cdot+0 \cdot\left({ }^{30} \mathrm{C}_0\right)^2 \\\\ & 2 \mathrm{~S}=30 \cdot\left({ }^{30} \mathrm{C}_0{ }^2+{ }^{30} \mathrm{C}_1{ }^2+\ldots \ldots . \cdot+{ }^{30} \mathrm{C}_{30}{ }^2\right) \\\\ & \mathrm{S}=15 \cdot{ }^{60} \mathrm{C}_{30}=15 \cdot \frac{60 !}{(30 !)^2} \\\\ & \frac{15 \cdot 10 !}{(30 !)^2}=\frac{\alpha \cdot 60 !}{(30 !)^2} \\\\ & \Rightarrow \alpha=15 \end{aligned}

Other Method : Given,

1\,.\,{\left( {{}^{30}{C_1}} \right)^2} + 2\,.\,{\left( {{}^{30}{C_2}} \right)^2} + 3\,.\,{\left( {{}^{30}{C_3}} \right)^2}\, + \,.....\, + \,30\,.\,{\left( {{}^{30}{C_{30}}} \right)^2}

= \sum\limits_{r = 1}^{30} {r\,.\,{{\left( {{}^{30}{C_r}} \right)}^2}}

= \sum\limits_{r = 1}^{30} {r\,.\,{}^{30}{C_r}\,.\,{}^{30}{C_r}}

= \sum\limits_{r = 1}^{30} {r\,.\,{{30} \over r}\,.\,{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}}

= 30\sum\limits_{r = 1}^{30} {{}^{29}{C_{r - 1}}\,.\,{}^{30}{C_r}}

= 30 \times

(Coefficient of

{x^{29}}

in

{(1 + x)^{59}}

)

= 30 \times \left( {{}^{59}{C_{29}}} \right)

= 30 \times {{60} \over {30}} \times {}^{59}{C_{29}} \times {{30} \over {60}}

= 30 \times {}^{60}{C_{30}} \times {{30} \over {60}}

= 15 \times {{60!} \over {30!\,30!}}

$\therefore$

\alpha = 15

Q64

The value of

\sum\limits_{r = 0}^{22} {{}^{22}{C_r}{}^{23}{C_r}}

is

A

{}^{44}{C_{23}}

B

{}^{45}{C_{23}}

C

{}^{44}{C_{22}}

D

{}^{45}{C_{24}}

Correct Answer

Option B

Solution

\sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,.\,{}^{23}{C_r}}

= \sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,{}^{23}{C_{23 - r}}}

[using

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

]

= {}^{22}{C_0}{}^{23}{C_{23}} + {}^{22}{C_1}{}^{23}{C_{22}}\, + \,...\, + \,{}^{22}{C_{21}}{}^{23}{C_2} + {}^{22}{C_{22}}{}^{23}{C_1}

We know,

{(1 + x)^{22}} = {}^{22}{C_0} + {}^{22}{C_1}x + {}^{22}{C_2}{x^2} + {}^{22}{C_3}{x^3}\, + \,...\, + \,{}^{22}{C_{22}}{x^{22}}

and

{(1 + x)^{23}} = {}^{23}{C_0} + {}^{23}{C_1}x + {}^{23}{C_2}{x^2}\, + \,...\, + \,{}^{23}{C_{22}}{x^{22}} + {}^{23}{C_{23}}{x^{23}}

Now coefficient of

{x^{23}}

in

{(1 + x)^{22}}{(1 + x)^{23}}

or

{(1 + x)^{45}}

= {}^{22}{C_0}{}^{23}{C_{23}} + {}^{22}{C_1}{}^{23}{C_{22}}\, + \,...\, + \,{}^{22}{C_{21}}\,.\,{}^{23}{C_2} + {}^{22}{C_{22}}\,.\,{}^{23}{C_1}

= \sum\limits_{r = 0}^{22} {{}^{22}{C_r}\,.\,{}^{23}{C_{23 - r}}}

$\therefore$ Coefficient of

{x^{23}}

in

{(1 + x)^{45}} = {}^{45}{C_{23}}

Q65

Let

\left(a+b x+c x^{2}\right)^{10}=\sum\limits_{i=0}^{20} p_{i} x^{i}, a, b, c \in \mathbb{N}

. If

p_{1}=20

and

p_{2}=210

, then

2(a+b+c)

is equal to :

A 15

B 8

C 6

D 12

Correct Answer

Option D

Solution

We are given that $\left(a+bx+cx^2\right)^{10} = \sum_{i=0}^{20} p_i x^i$ , and we are given that $p_1 = 20$ and $p_2 = 210$ .

We need to find the value of $2(a+b+c)$ .

Using the multinomial theorem, we can express the expansion of $(a + bx + cx^2)^{10}$ as follows:

\sum\limits_{k_1+k_2+k_3=10} {{10!} \over {{k_1}!{k_2}!{k_3}!}} a^{k_1} (bx)^{k_2} (cx^2)^{k_3}

Now we need to find the coefficients of $x^1$ and $x^2$ in the expansion: For $x^1$ term, we have:

k_2 = 1, k_1 = 9, k_3 = 0

So,

p_1 = {{10!} \over {9!1!0!}} a^9 b^1 = 10a^9 b

For $x^2$ term, there are two possibilities:

k_2 = 2, k_1 = 8, k_3 = 0 \quad \text{and} \quad k_2 = 0, k_1 = 9, k_3 = 1

So,

p_2 = {{10!} \over {8!2!0!}} a^8 b^2 + {{10!} \over {9!0!1!}} a^9 c = 45a^8 b^2 + 10a^9 c

Now we are given $p_1 = 20$ and $p_2 = 210$ . So,

10a^9 b = 20 \implies a^9 b = 2

and

45a^8 b^2 + 10a^9 c = 210

Now, divide the second equation by $a^8$ :

45b^2 + 10ac = 210

We know that $a^9 b = 2$ . Taking the $9^{th}$ root of both sides:

ab = \sqrt[9]{2}

Now, let $k = ab = \sqrt[9]{2}$ . We can rewrite the equation for $x^2$ term as:

45k^2 + 10k^9 = 210

From the equation $ab = k = \sqrt[9]{2}$ , we know that $a$ and $b$ are positive integers.

Thus, $k = 2$ (as both $a$ and $b$ must be factors of 2).

Now we have:

a+b = 2

and from the equation $a^9 b = 2$ , we get $a = 1, b = 2$ or vice versa.

Now we need to find the value of $c$ .

We can use the equation for the $x^2$ term again:

45a^8 b^2 + 10a^9 c = 210

Using $a=1$ and $b=2$ , we get:

45(1)^8 (2)^2 + 10(1)^9 c = 210 \implies 180 + 10c = 210 \implies c = 3

So, $a=1$ , $b=2$ , and $c=3$ . Now, we need to find the value of $2(a+b+c)$ :

2(a+b+c) = 2(1+2+3) = 2(6) = 12

Therefore, the answer is $\boxed{12}$ .

Q66

The sum, of the coefficients of the first 50 terms in the binomial expansion of

(1-x)^{100}

, is equal to

A

{ }^{99} \mathrm{C}_{49}

B

{ }^{101} \mathrm{C}_{50}

C

-{ }^{99} \mathrm{C}_{49}

D

-{ }^{101} \mathrm{C}_{50}

Correct Answer

Option C

Solution

\begin{aligned} & \left({ }^{100} C_0-{ }^{100} C_1+{ }^{100} C_2-\ldots . .{ }^{100} C_{49}\right)+{ }^{100} C_{50} \\\\ & +\left(-{ }^{100} C_{51}+{ }^{100} C_{52}-\ldots .+{ }^{100} C_{100}\right)=0 \\\\ & \lambda_1+{ }^{100} C_{50}+\lambda_2=0 \\\\ & \lambda_1=-\frac{1}{2}{ }^{100} C_{50} \quad\left(\because \lambda_1=\lambda_2\right) \\\\ & =-{ }^{99} C_{49} \end{aligned}

Q67

If

{ }^{2 n} C_{3}:{ }^{n} C_{3}=10: 1

, then the ratio

\left(n^{2}+3 n\right):\left(n^{2}-3 n+4\right)

is :

A

27: 11

B

2: 1

C

35: 16

D

65: 37

Correct Answer

Option B

Solution

\begin{aligned} & \text{We have, }{ }^{2 n} C_3:{ }^n C_3=10: 1 \\\\ & \Rightarrow \frac{{ }^{2 n} C_3}{{ }^n C_3}=\frac{10}{1} \\\\ & \Rightarrow \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=\frac{10}{1} \\\\ & \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{(n)(n-1)(n-2)}=\frac{10}{1} \\\\ & \Rightarrow 4 n^2-6 n+2=5\left(n^2-3 n+2\right) \\\\ & \Rightarrow n^2-9 n+8=0 \\\\ & \Rightarrow n^2-8 n-n+8=0 \\\\ & \Rightarrow n(n-8)-1(n-8)=0 \\\\ & \Rightarrow (n-8)(n-1)=0 \\\\ & \Rightarrow n=8(n=1 \text{ not valid }) \end{aligned}

\therefore \frac{n^2+3 n}{n^2-3 n+4}=\frac{88}{44}=\frac{2}{1}=2: 1

Q68

The sum of the coefficients of three consecutive terms in the binomial expansion of

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

, which are in the ratio

1: 3: 5

, is equal to :

A 63

B 92

C 25

D 41

Correct Answer

Option A

Solution

The problem asks for the sum of the coefficients of three consecutive terms in the binomial expansion of $(1+x)^{n+2}$ , which are in the ratio 1 : 3 : 5.

Given that the ratios of the coefficients are 1:3:5, we let the terms be $T_r$ , $T_{r+1}$ , and $T_{r+2}$ .

The coefficients of these terms are ${ }^{n+2} C_{r-1}$ , ${ }^{n+2} C_{r}$ , and ${ }^{n+2} C_{r+1}$ , respectively.

\begin{aligned} & \frac{T_{r+1}}{T_r}=\frac{{ }^{n+2} C_r}{{ }^{n+2} C_{r-1}}=\frac{n+2-r+1}{r}=\frac{n+3-r}{r}=3 \\\\ & n-4 r+3=0 ......(1) \\\\ & \frac{T_{r+2}}{T_{r+1}}=\frac{{ }^{n+2} C_{r+1}}{{ }^{n+2} C_r}=\frac{(n+2)-(r+1)+1}{r+1}=\frac{n-r+2}{r+1}=\frac{5}{3} \\\\ & 3 n-8 r+1=0 ......(2) \end{aligned}

By solving (1) and (2), we get $\Rightarrow n=5, r=2$

\begin{aligned} T_r+T_{r+1}+T_{r+2} & ={ }^7 C_1+{ }^7 C_2+{ }^7 C_3 \\\\ & =7+21+35=63 \end{aligned}

Q69

If the

1011^{\text{th }}

term from the end in the binominal expansion of

\left(\dfrac{4 x}{5}-\dfrac{5}{2 x}\right)^{2022}

is 1024 times

1011^{\text{th }}

R term from the beginning, then

|x|

is equal to

A

\dfrac{5}{16}

B 8

C 12

D 15

Correct Answer

Option A

Solution

$\mathrm{T}_{1011}$ from beginning $=\mathrm{T}_{1010+1}$

={ }^{2022} \mathrm{C}_{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}

$\mathrm{T}_{1011}$ from end

={ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010}

\begin{aligned} & \text{ Given : }{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1012}\left(\frac{4 \mathrm{x}}{5}\right)^{1010} \\\\ & =2^{10} \cdot{ }^{2022} \mathrm{C}_{1010}\left(\frac{-5}{2 \mathrm{x}}\right)^{1010}\left(\frac{4 \mathrm{x}}{5}\right)^{1012} \end{aligned}

\begin{aligned} &\Rightarrow \left(\frac{-5}{2 x}\right)^2=2^{10}\left(\frac{4 x}{5}\right)^2 \\\\ &\Rightarrow x^4=\frac{5^4}{2^{16}} \\\\ & \Rightarrow |x|=\frac{5}{16} \end{aligned}

Q70

Let the number

(22)^{2022}+(2022)^{22}

leave the remainder

\alpha

when divided by 3 and

\beta

when divided by 7. Then

\left(\alpha^{2}+\beta^{2}\right)

is equal to :

A 13

B 10

C 20

D 5

Correct Answer

Option D

Solution

We have, $(22)^{2022}+(2022)^{22}$ As 2022 is completely divisible by 3 So, $(2022)^{22}$ is also divisible by 3 $(22)^{2022}=(21+1)^{2022}=(3 \times 7+1)^{2022}=7 m+1$ $\Rightarrow(22)^{2022}$ leave a remainder 1 , when divisible by 3 . $\therefore(22)^{2022}+(2022)^{22}$ leave a remainder when divisible by 3 $\therefore \alpha=1$

\begin{aligned} (22)^{2022}+(2022)^{22} & =(21+1)^{2022}+(2023-1)^{22} \\\\ & =7 K+1+7 \mu+1=7(K+\mu)+2 \end{aligned}

$\Rightarrow(22)^{2022}+(2022)^{22}$ leave a remainder 2 when divisible by 7

\begin{aligned} & \therefore \beta=2 \\\\ & \text{ Hence, } \alpha^2+\beta^2=1^2+2^2=5 \end{aligned}