SATHEE: Binomial Theorem

Binomial Theorem

Short Answer Type Questions

1. Find the term independent of $x$ , where $x \neq 0$ ,

in the expansion of $(\frac{3 x^{2}}{2} - \frac{1}{3 x})^{15}$ .

Thinking Process

The general term in the expansion of $(x - a)^{n}$ i.e., $T_{r + 1} =^{n} C_{r} (x)^{n - r} (- a)^{r}$ . For the term independent of $x$ , put $n - r = 0$ , then we get the value of $r$ .

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Solution

Given expansion is $(\frac{3 x^{2}}{2} - \frac{1}{3 x})^{15}$

Let $T_{r + 1}$ term is the general term.

Then,

$\begin{aligned} T_{r + 1} & =^{15} C_{r} (\frac{3 x^{2}}{2})^{15 - r} (- \frac{1}{3 x})^{r} \\ =^{15} C_{r} 3^{15 - r} x^{30 - 2 r} 2^{r - 15} (- 1)^{r} \cdot 3^{- r} \cdot x^{- r} \\ =^{15} C_{r} (- 1)^{r} 3^{15 - 2 r} 2^{r - 15} x^{30 - 3 r} \end{aligned}$

For independent of $x$ ,

$\begin{aligned} 30 - 3 r & = 0 \\ 3 r & = 30 \Rightarrow r = 10 \\ ∵ T_{r + 1} & = T_{10 + 1} = 11 th term is independent of x . \\ ∴ T_{10 + 1} & =^{15} C_{10} (- 1)^{10} 3^{15 - 20} 2^{10 - 15} \\ =^{15} C_{10} 3^{- 5} 2^{- 5} \\ =^{15} C_{10} (6)^{- 5} \\ =^{15} C_{10} (\frac{1}{6})^{5} \end{aligned}$

2. If the term free from $x$ in the expansion of $(\sqrt{x} - \frac{k}{x^{2}})^{10}$ is 405 , then find the value of $k$ .

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Solution

Given expansion is $(\sqrt{x} - \frac{k}{x^{2}})^{10}$ .

Let $T_{r + 1}$ is the general term.

Then,

$\begin{aligned} T_{r + 1} & =^{10} C_{r} (\sqrt{x})^{10 - r} (\frac{- k}{x^{2}})^{r} \\ =^{10} C_{r} (x)^{\frac{1}{2} (10 - r)} (- k)^{r} \cdot x^{- 2 r} \\ =^{10} C_{r} x^{5 - \frac{r}{2}} (- k)^{r} \cdot x^{- 2 r} \\ =^{10} C_{r} x^{5 - \frac{r}{2} - 2 r} (- k)^{r} \\ =^{10} C_{r} x^{\frac{10 - 5 r}{2}} (- k)^{r} \end{aligned}$

For free from $x, \frac{10 - 5 r}{2} = 0$

$\Rightarrow 10 - 5 r = 0 \Rightarrow r = 2$

Since, $T_{2 + 1} = T_{3}$ is free from $x$ .

$\begin{aligned} ∴ & T_{2 + 1} & =^{10} C_{2} (- k)^{2} = 405 \\ \Rightarrow & \frac{10 \times 9 \times 8!}{2! \times 8!} (- k)^{2} & = 405 \\ \Rightarrow 45 k^{2} & = 405 \Rightarrow & k^{2} = \frac{405}{45} = 9 \\ ∴ & k & = \pm 3 \end{aligned}$

3. Find the coefficient of $x$ in the expansion of $(1 - 3 x + 7 x^{2}) (1 - x)^{16}$ .

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Solution

Given, expansion $= (1 - 3 x + 7 x^{2}) (1 - x)^{16}$ .

$\begin{aligned} = (1 - 3 x + 7 x^{2}) (^{16} C_{0} 1^{16} -^{16} C_{1} 1^{15} x^{1} +^{16} C_{2} 1^{14} x^{2} + \dots +^{16} C_{16} x^{16}) \\ = (1 - 3 x + 7 x^{2}) (1 - 16 x + 120 x^{2} + \dots) \end{aligned}$

$∴$ Coefficient of $x = - 3 - 16 = - 19$

4. Find the term independent of $x$ in the expansion of $(3 x - \frac{2}{x^{2}})^{15}$ .

Thinking Process

The general term in the expansion of $(x - a)^{n}$ i.e., $T_{r + 1} =^{n} C_{r} (x)^{n - r} (- a)^{r}$ .

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Solution

Given expansion is $(3 x - \frac{2}{x^{2}})^{15}$ .

Let $T_{r + 1}$ is the general term.

$\begin{aligned} ∴ T_{r + 1} & =^{15} C_{r} (3 x)^{15 - r} \frac{- 2}{x^{2}} =^{r}^{15} C_{r} (3 x)^{15 - r} (- 2)^{r} x^{- 2 r} \\ =^{15} C_{r} 3^{15 - r} x^{15 - 3 r} (- 2)^{r} \end{aligned}$

For independent of $x, 15 - 3 r = 0 \Rightarrow r = 5$

Since, $T_{5 + 1} = T_{6}$ is independent of $x$ .

$\begin{aligned} ∴ T_{5 + 1} & =^{15} C_{5} 3^{15 - 5} (- 2)^{5} \\ = - \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10!}{5 \times 4 \times 3 \times 2 \times 1 \times 10!} \cdot 3^{10} \cdot 2^{5} \\ = - 3003 \cdot 3^{10} \cdot 2^{5} \end{aligned}$

5. Find the middle term (terms) in the expansion of

(i) $(\frac{x}{a} - \frac{a}{x})^{10}$

(ii) $(3 x - \frac{x^{3}}{6})^{9}$

Thinking Process

In the expansion of $(a + b)^{n}$ , if $n$ is even, then this expansion has only one middle term i.e., $(\frac{n}{2} + 1)$ th term is the middle term and if $n$ is odd, then this expansion has two middle terms i.e., $(\frac{n + 1}{2})$ thand $(\frac{n + 1}{2} + 1)$ thare two middle terms.

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Solution

(i) Given expansion is $(\frac{x}{a} - \frac{a}{x})^{10}$

Here, the power of Binomial i.e., $n = 10$ is even.

Since, it has one middle term $(\frac{10}{2} + 1)$ th term i.e., 6th term.

$\begin{aligned} ∴ T_{6} & = T_{5 + 1} =^{10} C_{5} (\frac{x}{a})^{10 - 5} (\frac{- a}{x})^{5} \\ = -^{5} C_{5} (\frac{x}{a})^{5} (\frac{a}{x})^{5} \\ = - \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 5 \times 4 \times 3 \times 2 \times 1} (\frac{x}{a})^{5} (\frac{x}{a})^{- 5} \\ = - 9 \times 4 \times 7 = - 252 \end{aligned}$

(ii) Given expansion is $(3 x - \frac{x^{3}}{6})^{9}$ .

Here, $n = 9$ [odd]

Since, the Binomial expansion has two middle terms i.e., $(\frac{9 + 1}{2})$ th and $(\frac{9 + 1}{2} + 1)$ th i.e., 5 th term and 6 th term.

$\begin{aligned} ∴ T_{5} & = T_{(4 + 1)} =^{9} C_{4} (3 x)^{9 - 4} (- \frac{x^{3}}{6})^{4} \\ = \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} 3^{5} x^{5} x^{12} 6^{- 4} \\ = \frac{7 \times 6 \times 3 \times 3^{1}}{2^{4}} x^{17} = \frac{189}{8} x^{17} \end{aligned}$

$\begin{aligned} ∴ T_{6} & = T_{5 + 1} =^{9} C_{5} (3 x)^{9 - 5} (- \frac{x^{3}}{6})^{5} \\ = - \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4 \times 3 \times 2 \times 1} \cdot 3^{4} \cdot x^{4} \cdot x^{15} \cdot 6^{- 5} \\ = \frac{- 21 \times 6}{3 \times 2^{5}} x^{19} = \frac{- 21}{16} x^{19} \end{aligned}$

6. Find the coefficient of $x^{15}$ in the expansion of $(x - x^{2})^{10}$ .

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Solution

Given expansion is $(x - x^{2})^{10}$ .

Let the term $T_{r + 1}$ is the general term.

$\begin{aligned} ∴ T_{r + 1} & =^{10} C_{r} x^{10 - r} (- x^{2})^{r} \\ = (- 1)^{r} \cdot^{10} C_{r} \cdot x^{10 - r} \cdot x^{2 r} \\ = (- 1)^{1}^{10} C_{r} x^{10 + r} \end{aligned}$

For the coefficient of $x^{15}$ ,

$\begin{aligned} 10 + r & = 15 \Rightarrow r = 5 \\ T_{5 + 1} & = (- 1)^{5}^{10} C_{5} x^{15} \\ ∴ Coefficient of x^{15} & = - 1 \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5 \times 4 \times 3 \times 2 \times 1 \times 5!} \\ = - 3 \times 2 \times 7 \times 6 = - 252 \end{aligned}$

7. Find the coefficient of $\frac{1}{x^{17}}$ in the expansion of $(x^{4} - \frac{1}{x^{3}})^{15}$ .

Thinking Process

In this type of questions, first of all find the general terms, in the expansion $(x - y)^{n}$ using the formula $T_{r + 1} =^{n} C_{r} x^{n - r} (- y)^{r}$ and then put $n - r$ equal to the required power of $x$ of which coefficient is to be find out.

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Solution

Given expansion is $(x^{4} - \frac{1}{x^{3}})^{15}$ .

Let the term $T_{r + 1}$ contains the coefficient of $\frac{1}{x^{17}}$ i.e., $x^{- 17}$ .

$∴ \begin{aligned} T_{r + 1} & =^{15} C_{r} (x^{4})^{15 - r} (- \frac{1}{x^{3}})^{r} \\ =^{15} C_{r} x^{60 - 4 r} (- 1)^{r} x^{- 3 r} \\ =^{15} C_{r} x^{60 - 7 r} (- 1)^{r} \end{aligned}$

For the coefficient $x^{- 17}$ ,

$\begin{aligned} 60 - 7 r = - 17 \\ \Rightarrow 7 r = 77 \Rightarrow r = 11 \\ \Rightarrow T_{11 + 1} =^{15} C_{11} x^{60 - 77} (- 1)^{11} \\ ∴ Coefficient of x^{- 17} = \frac{- 15 \times 14 \times 13 \times 12 \times 11!}{11! \times 4 \times 3 \times 2 \times 1} \\ = - 15 \times 7 \times 13 = - 1365 \end{aligned}$

8. Find the sixth term of the expansion $(y^{1 / 2} + x^{1 / 3})^{n}$ , if the Binomial coefficient of the third term from the end is 45 .

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Solution

Given expansion is $(y^{1 / 2} + x^{1 / 3})^{n}$ .

The sixth term of this expansion is

$T_{6} = T_{5 + 1} =^{n} C_{5} (y^{1 / 2})^{n - 5} (x^{1 / 3})^{5} \dots (i)$

Now, given that the Binomial coefficient of the third term from the end is 45 .

We know that, Binomial coefficient of third term from the end =Binomial coefficient of third term from the begining $=^{n} C_{2}$

$\begin{aligned} ∵^{n} C_{2} = 45 \\ \Rightarrow \frac{n (n - 1) (n - 2)!}{2! (n - 2)!} = 45 \\ \Rightarrow n (n - 1) = 90 \\ \Rightarrow n^{2} - n - 90 = 0 \\ \Rightarrow n^{2} - 10 n + 9 n - 90 = 0 \\ \Rightarrow n (n - 10) + 9 (n - 10) = 0 \\ \Rightarrow (n - 10) (n + 9) = 0 \\ \Rightarrow (n + 9) = 0 or (n - 10) = 0 \\ ∴ n = 10 [∵ n \neq - 9] \end{aligned}$

From Eq. (i),

$T_{6} =^{10} C_{5} y^{5 / 2} x^{5 / 3} = 252 y^{5 / 2} \cdot x^{5 / 3}$

9. Find the value of $r$ , if the coefficients of $(2 r + 4)$ th and $(r - 2)$ th terms in the expansion of $(1 + x)^{18}$ are equal.

Thinking Process

Coefficient of $(r + 1)$ th term in the expansion of $(1 + x)^{n}$ is $^{n} C_{r}$ . Use this formula to solve the above problem.

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Solution

Given expansion is $(1 + x)^{18}$ .

Now, $(2 r + 4)$ th term i.e., $T_{2 r + 3 + 1}$ .

$\begin{aligned} ∴ T_{2 r + 3 + 1} & =^{18} C_{2 r + 3} (1)^{18 - 2 r - 3} (x)^{2 r + 3} \\ =^{18} C_{2 r + 3} x^{2 r + 3} \end{aligned}$

Now, $(r - 2)$ th term i.e., $T_{r - 3 + 1}$ .

$∴ T_{r - 3 + 1} =^{18} C_{r - 3} x^{r - 3}$

${As,}^{18} C_{2 r + 3} =^{18} C_{r - 3}$ $[∵^{n} C_{x} =^{n} C_{y} \Rightarrow x + y = n]$

$\Rightarrow 2 r + 3 + r - 3 = 18$

$\Rightarrow 3 r = 18$

$∴ r = 6$

10. If the coefficient of second, third and fourth terms in the expansion of $(1 + x)^{2 n}$ are in AP, then show that $2 n^{2} - 9 n + 7 = 0$ .

Thinking Process

In the expansion of $(x + y)^{n}$ , the coefficient of $(r + 1)$ th term is $^{n} C_{r}$ . Use this formula to get the required coefficient. If $a, b$ and $c$ are in $A P$ , then $2 b = a + c$ .

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Solution

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

Now, coefficient of 2nd term $=^{2 n} C_{1}$

Coefficient of 3rd term $=^{2 n} C_{2}$

Coefficient of 4 th term $=^{2 n} C_{3}$

Given that, $^{2 n} C_{1},^{2 n} C_{2}$ and $^{2 n} C_{3}$ are in AP.

Then, $2 \cdot^{2 n} C_{2} =^{2 n} C_{1} +^{2 n} C_{3}$

$\Rightarrow 2 [\frac{2 n (2 n - 1) (2 n - 2)!}{2 \times 1 \times (2 n - 2)!}] = \frac{2 n (2 n - 1)!}{(2 n - 1)!} + \frac{2 n (2 n - 1) (2 n - 2) (2 n - 3)!}{3! (2 n - 3)!}$

$\Rightarrow n (2 n - 1) = n + \frac{n (2 n - 1) (2 n - 2)}{6}$

$\Rightarrow n (12 n - 6) = n (6 + 4 n^{2} - 4 n - 2 n + 2)$

$\Rightarrow 12 n - 6 = (4 n^{2} - 6 n + 8)$

$\Rightarrow 6 (2 n - 1) = 2 (2 n^{2} - 3 n + 4)$

$\Rightarrow 3 (2 n - 1) = 2 n^{2} - 3 n + 4$

$\Rightarrow 2 n^{2} - 3 n + 4 - 6 n + 3 = 0$

$\Rightarrow 2 n^{2} - 9 n + 7 = 0$

11. Find the coefficient of $x^{4}$ in the expansion of $(1 + x + x^{2} + x^{3})^{11}$ .

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Solution

Given, expansion $= (1 + x + x^{2} + x^{3})^{11} = [(1 + x) + x^{2} (1 + x)]^{11}$

$= [(1 + x) (1 + x^{2})]^{11} = (1 + x)^{11} \cdot (1 + x^{2})^{11}$

Now, above expansion becomes

$\begin{aligned} = (^{11} C_{0} +^{11} C_{1} x +^{11} C_{2} x^{2} +^{11} C_{3} x^{3} +^{11} C_{4} x^{4} + \dots) (^{11} C_{0} +^{11} C_{1} x^{2} +^{11} C_{2} x^{4} + \dots) \\ = (1 + 11 x + 55 x^{2} + 165 x^{3} + 330 x^{4} + \dots) (1 + 11 x^{2} + 55 x^{4} + \dots) \end{aligned}$

$∴$ Coefficient of $x^{4} = 55 + 605 + 330 = 990$

Long Answer Type Questions

12. If $p$ is a real number and the middle term in the expansion of $(\frac{p}{2} + 2)^{8}$ is 1120, then find the value of $p$ .

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Solution

Given expansion is $(\frac{p}{2} + 2)^{8}$ .

Here, $n = 8$ [even]

Since, this Binomial expansion has only one middle term i.e., $(\frac{8}{2} + 1)$ th $= 5$ th term

$T_{5} = T_{4 + 1} =^{8} C_{4} (\frac{p}{2})^{8 - 4} \cdot 2^{4}$

$\Rightarrow 1120 =^{8} C_{4} p^{4} \cdot 2^{- 4} 2^{4}$

$\Rightarrow 1120 = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4 \times 3 \times 2 \times 1} p^{4}$

$\begin{matrix} \Rightarrow & 1120 = 7 \times 2 \times 5 \times p^{4} \\ \Rightarrow & p^{4} = \frac{1120}{70} = 16 \Rightarrow p^{4} = 2^{4} \\ \Rightarrow & p^{2} = 4 \Rightarrow p = \pm 2 \end{matrix}$

13. Show that the middle term in the expansion of $(x - \frac{1}{x})^{2 n}$ is

$\frac{1 \times 3 \times 5 \times \dots \times (2 n - 1)}{n!} \times (- 2)^{n}$

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Solution

Given, expansion is $(x - \frac{1}{x})^{2 n}$ . This Binomial expansion has even power. So, this has one middle term.

i.e., $\begin{aligned} (\frac{2 n}{2} + 1) & th term = (n + 1) th term \\ T_{n + 1} & =^{2 n} C_{n} (x)^{2 n - n} (- \frac{1}{x})^{n} =^{2 n} C_{n} x^{n} (- 1)^{n} x^{- n} \\ =^{2 n} C_{n} (- 1)^{n} = (- 1)^{n} \frac{(2 n)!}{n! n!} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \dots (2 n - 1) (2 n)}{n! n!} (- 1)^{n} \\ = \frac{1 \cdot 3 \cdot 5 \dots (2 n - 1) \cdot 2 \cdot 4 \cdot 6 \dots (2 n)}{12 \cdot 3 \cdot \dots n (n!)} (- 1)^{n} \\ = \frac{1 \cdot 3 \cdot 5 \dots (2 n - 1) \cdot 2^{n} (1 \cdot 2 \cdot 3 \dots n) (- 1)^{n}}{(1 \cdot 2 \cdot 3 \dots n) (n!)} \\ = \frac{[1 \cdot 3 \cdot 5 \dots (2 n - 1)]}{n!} (- 2)^{n} \end{aligned}$ Hence proved.

14. Find $n$ in the Binomial $(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})^{n}$ , if the ratio of 7th term from the beginning to the 7th term from the end is $\frac{1}{6}$ .

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Solution

Here, the Binomial expansion is $(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}})^{n}$ .

Now, 7th term from beginning $T_{7} = T_{6 + 1} =^{n} C_{6} (\sqrt[3]{2})^{n - 6} (\frac{1}{\sqrt[3]{3}})^{6} \dots (i)$

and 7 th term from end i.e., $T_{7}$ from the beginning of $(\frac{1}{\sqrt[3]{3}} + \sqrt[3]{2})^{n}$

i.e., $T_{7} =^{n} C_{6} (\frac{1}{\sqrt[3]{3}})^{n - 6} (\sqrt[3]{2})^{6}$

Given that, $\frac{^{n} C_{6} (\sqrt[3]{2})^{n - 6} (\frac{1}{\sqrt[3]{3}})^{6}}{^{n} C_{6} (\frac{1}{\sqrt[3]{3}})^{n - 6} (\sqrt[3]{2})^{6}} = \frac{1}{6} \Rightarrow \frac{2^{\frac{n - 6}{3}} \cdot 3^{- 6 / 3}}{3^{- \frac{n - 6}{3}} \cdot 2^{6 / 3}} = \frac{1}{6}$

$\Rightarrow (2^{\frac{n - 6}{3}} \cdot 2^{\frac{- 6}{3}}) (3^{\frac{- 6}{3}} \cdot 3^{\frac{(n - 6)}{3}}) = 6^{- 1}$

$\begin{aligned} \Rightarrow 2^{\frac{n - 6}{3} - \frac{6}{3}} \cdot 3^{\frac{n - 6}{3} - \frac{6}{3}} = 6^{- 1} \Rightarrow (2 \cdot 3)^{\frac{n}{3} - 4} = 6^{- 1} \\ \Rightarrow \frac{n}{3} - 4 = - 1 \Rightarrow \frac{n}{3} = 3 \\ ∴ n = 9 \end{aligned}$

15. In the expansion of $(x + a)^{n}$ , if the sum of odd terms is denoted by 0 and the sum of even term by $E$ . Then, prove that

(i) $0^{2} - E^{2} = (x^{2} - a^{2})^{n}$ .

(ii) $40 E = (x + a)^{2 n} - (x - a)^{2 n}$ .

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Solution

(i) Given expansion is $(x + a)^{n}$ .

$∴ (x + a)^{n} =^{n} C_{0} x^{n} a^{0} +^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} +^{n} C_{3} x^{n - 3} a^{3} + \dots +^{n} C_{n} a^{n}$

Now, sum of odd terms

i.e.,

$O =^{n} C_{0} x^{n} +^{n} C_{2} x^{n - 2} a^{2} + \dots$

and sum of even terms

i.e.,

$E =^{n} C_{1} x^{n - 1} a +^{n} C_{3} x^{n - 3} a^{3} + \dots$

$\begin{aligned} ∵ (x + a)^{n} = O + E \\ Similarly, (x - a)^{n} = O - E \\ ∴ (O + E) (O - E) = (x + a)^{n} (x - a)^{n} [on multiplying Eqs. (i) and (ii)] \\ \Rightarrow O^{2} - E^{2} = (x^{2} - a^{2})^{n} \end{aligned}$

(ii) $4 O E = (O + E)^{2} - (O - E)^{2} = [(x + a)^{n}]^{2} - [(x - a)^{n}]^{2}$

[from Eqs. (i) and (ii)]

$= (x + a)^{2 n} - (x - a)^{2 n}$

Hence proved.

16. If $x^{p}$ occurs in the expansion of $x^{2} + {\frac{1}{x}}^{2 n}$ , then prove that its coefficient is $\frac{2 n!}{\frac{(4 n - p)!}{3!} \frac{(2 n + p)!}{3!}}$ .

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Solution

Given expansion is $x^{2} + {\frac{1}{x}}^{2 n}$ .

Let $x^{p}$ occur in the expansion of $x^{2} + {\frac{1}{x}}^{2 n}$ .

$\begin{aligned} T_{r + 1} & =^{2 n} C_{r} (x^{2})^{2 n - r} \frac{1}{x} \\ =^{2 n} C_{r} x^{4 n - 2 r} x^{- r} =^{2 n} C_{r} x^{4 n - 3 r} \\ Let 4 n - 3 r & = p \\ \Rightarrow 3 r & = 4 n - p \Rightarrow r = \frac{4 n - p}{3} \\ ∴ Coefficient of x^{p} & =^{2 n} C_{r} = \frac{(2 n)!}{r! (2 n - r)!} = \frac{(2 n)!}{\frac{4 n - p}{3}! 2 n - \frac{4 n - p}{3}!} \\ = \frac{(2 n)!}{\frac{4 n - p}{3}! \frac{6 n - 4 n + p}{3}!} = \frac{(2 n)!}{\frac{4 n - p}{3}! \frac{2 n + p}{3}!} \end{aligned}$

17. Find the term independent of $x$ in the expansion of

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

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Solution

Given expansion is $(1 + x + 2 x^{3}) \frac{3}{2} x^{2} - {\frac{1}{3 x}}^{9}$ .

Now, consider $\frac{3}{2} x^{2} - {\frac{1}{3 x}}^{9}$

$\begin{aligned} T_{r + 1} & =^{9} C_{r} \frac{3}{2} x^{2} - \frac{1}{3 x} \\ =^{9} C_{r} {\frac{3}{2}}^{9 - r} x^{18 - 2 r} - {\frac{1}{3}}^{r} x^{- r} =^{9} C_{r} {\frac{3}{2}}^{9 - r} - {\frac{1}{3}}^{r} x^{18 - 3 r} \end{aligned}$

Hence, the general term in the expansion of $(1 + x + 2 x^{3}) \frac{3}{2} x^{2} - {\frac{1}{3 x}}^{9}$

$=^{9} C_{r} {\frac{3}{2}}^{9 - r} - {\frac{1}{3}}^{r} x^{18 - 3 r} +^{9} C_{r} {\frac{3}{2}}^{9 - r} - {\frac{1}{3}}^{r} x^{19 - 3 r} + 2 \cdot^{9} C_{r} {\frac{3}{2}}^{9 - r} - {\frac{1}{3}}^{r} x^{21 - 3 r}$

For term independent of $x$ , putting $18 - 3 r = 0, 19 - 3 r = 0$ and $21 - 3 r = 0$ , we get

$r = 6, r = 19 / 3, r = 7$

Since, the possible value of $r$ are 6 and 7 .

Hence, second term is not independent of $x$ .

$∴$ The term independent of $x$ is $^{9} C_{6} {\frac{3}{2}}^{9 - 6} - {\frac{1}{3}}^{6} + 2 \cdot^{9} C_{7} {\frac{3}{2}}^{9 - 7} - {\frac{1}{3}}^{7}$

$\begin{aligned} = \frac{9 \times 8 \times 7 \times 6!}{6! \times 3 \times 2} \cdot \frac{3^{3}}{2^{3}} \cdot \frac{1}{3^{6}} - 2 \cdot \frac{9 \times 8 \times 7!}{7! \times 2 \times 1} \cdot \frac{3^{2}}{2^{2}} \cdot \frac{1}{3^{7}} \\ = \frac{84}{8} \cdot \frac{1}{3^{3}} - \frac{36}{4} \cdot \frac{2}{3^{5}} = \frac{7}{18} - \frac{2}{27} = \frac{21 - 4}{54} = \frac{17}{54} \end{aligned}$

Objective Type Questions

18. The total number of terms in the expansion of $(x + a)^{100} + (x - a)^{100}$ after simplification is

(a) 50

(b) 202

(d) None of these

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Solution

Total number of terms is 102 in the expansion of $(x + a)^{100} + (x - a)^{100}$

50 terms of $(x + a)^{100}$ cancel out 50 terms of $(x - a)^{100} .51$ terms of $(x + a)^{100}$ get added to the 51 terms of $(x - a)^{100}$ .

Alternate Method

$\begin{aligned} (x + a)^{100} + (x - a)^{100} & =^{100} C_{0} x^{100} +^{100} C_{1} x^{99} a + \dots +^{100} C_{100} a^{100} \\ +^{100} C_{0} x^{100} -^{100} C_{1} x^{99} a + \dots +^{100} C_{100} a^{100} \\ = 2 \underset{51 terms}{\underset{⏟}{.^{100} C_{0} x^{100} +^{100} C_{2} x^{98} a^{2} + \dots +^{100} C_{100} a^{100}]}} \end{aligned}$

19. If the integers $r > 1, n > 2$ and coefficients of $(3 r)$ th and $(r + 2)$ nd terms in the Binomial expansion of $(1 + x)^{2 n}$ are equal, then

(a) $n = 2 r$

(b) $n = 3 r$

(d) None of these

Thinking Process

In the expansion of $(x + y)^{n}$ , the coefficient of $(r + 1)$ th term is $^{n} C_{r}$ .

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Solution

(a) Given that, $r > 1, n > 2$ and the coefficients of $(3 r)$ th and $(r + 2)$ th term are equal in the expansion of $(1 + x)^{2 n}$ .

$\begin{aligned} Then, T_{3 r} = T_{3 r - 1 + 1} =^{2 n} C_{3 r - 1} x^{3 r - 1} \\ and T_{r + 2} = T_{r + 1 + 1} =^{2 n} C_{r + 1} x^{r + 1} \\ Given,^{2 n} C_{3 r - 1} =^{2 n} C_{r + 1} [∵^{n} C_{x} =^{n} C_{y} \Rightarrow x + y = n] \\ \Rightarrow 3 r - 1 + r + 1 = 2 n \\ \Rightarrow 4 r = 2 n \Rightarrow n = \frac{4 r}{2} \\ ∴ n = 2 r \end{aligned}$

20. The two successive terms in the expansion of $(1 + x)^{24}$ whose coefficients are in the ratio $1 : 4$ are

(a) 3rd and 4th

(b) 4th and 5th

(d) 6th and 7th

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Solution

$∴$

and

$T_{r + 1} =^{24} C_{r} x^{r}$

$T_{r + 2} =^{24} C_{r + 1} x^{r + 1}$

$\frac{^{24} C_{r}}{^{24} C_{r + 1}} = \frac{1}{4}$

Given that,

$\Rightarrow \frac{\frac{(24)!}{r! (24 - r)!}}{(24)!} = \frac{1}{4}$

$\Rightarrow \frac{(r + 1)! (24 - r - 1)!}{(r + 1) r (23 - r)!} = \frac{1}{4}$

$\Rightarrow \frac{(r + 124 - r) (23 - r)!}{r 4 - r} = \frac{1}{4} \Rightarrow 4 r + 4 = 24 - r$

$\Rightarrow 5 r = 20 \Rightarrow r = 4$

$∴ T_{4 + 1} = T_{5} and T_{4 + 2} = T_{6}$

Hence, 5th and 6th terms.

21. The coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n}$ and $(1 + x)^{2 n - 1}$ are in the ratio

(a) $1 : 2$

(b) $1 : 3$

(d) $2 : 1$

Show Answer

Solution

(d) $∵$ Coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n} =^{2 n} C_{n}$ and coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n - 1} =^{2 n - 1} C_{n}$

$\frac{^{2 n} C_{n}}{^{2 n - 1} C_{n}} = \frac{\frac{(2 n)!}{n! n!}}{\frac{(2 n - 1)!}{n! (n - 1)!}}$

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

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

$= \frac{2 n}{n} = \frac{2}{1} = 2 : 1$

22. If the coefficients of $2 n d$ , 3rd and the 4th terms in the expansion of $(1 + x)^{n}$ are in AP, then the value of $n$ is

(a) 2

(b) 7

(d) 14

Show Answer

Solution

(b) The expansion of $(1 + x)^{n}$ is $^{n} C_{0} +^{n} C_{1} x +^{n} C_{2} x^{2} +^{n} C_{3} x^{3} + \dots +^{n} C_{n} x^{n}$

$∴$ Coefficient of 2 nd term $=^{n} C_{1}$ ,

Coefficient of 3rd term $=^{n} C_{2}$

and coefficient of 4 th term $=^{n} C_{3}$

Given that, $^{n} C_{1},^{n} C_{2}$ and $^{n} C_{3}$ are in AP

$\begin{matrix} ∴ & 2^{n} C_{2} =^{n} C_{1} +^{n} C_{3} \\ \Rightarrow & 2 \frac{(n)!}{(n - 2)! 2!} = \frac{(n)!}{(n - 1)!} + \frac{(n)!}{3! (n - 3)!} \\ \Rightarrow & \frac{2 \cdot n (n - 1) (n - 2)!}{(n - 2)! 2!} = \frac{n (n - 1)!}{(n - 1)!} + \frac{n (n - 1) (n - 2) (n - 3)!}{3 \cdot 2 \cdot 1 (n - 3)!} \\ \Rightarrow & n (n - 1) = n + \frac{n (n - 1) (n - 2)}{6} \\ \Rightarrow & 6 n - 6 = 6 + n^{2} - 3 n + 2 \\ \Rightarrow & n^{2} - 9 n + 14 = 0 \\ \Rightarrow & n (n - 7) - 2 (n - 7) = 0 \\ \Rightarrow & (n - 7) (n - 2) = 0 \\ \Rightarrow & n = 2 or n = 7 \end{matrix}$

Since, $n = 2$ is not possible.

$∴ n = 7$

23. If $A$ and $B$ are coefficient of $x^{n}$ in the expansions of $(1 + x)^{2 n}$ and $(1 + x)^{2 n - 1}$ respectively, then $\frac{A}{B}$ equals to

(a) 1

(b) 2

(d) $\frac{1}{n}$

Show Answer

Solution

(b) Since, the coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n}$ is $^{2 n} C_{n}$ .

$∴ A =^{2 n} C_{n}$

Now, the coefficient of $x^{n}$ in the expansion of $(1 + x)^{2 n - 1}$ is $^{2 n - 1} C_{n}$ .

$∴ B =^{2 n - 1} C_{n}$

Now,

$\frac{A}{B} = \frac{^{2 n} C_{n}}{^{2 n - 1} C_{n}} = \frac{2}{1} = 2$

Same as solution No. 21.

24. If the middle term of $\frac{1}{x} + x \sin x$ is equal to $7 \frac{7}{8}$ , then the value of $x$ is

(a) $2 n π + \frac{π}{6}$

(b) $n π + \frac{π}{6}$

(d) $n π + (- 1)^{n} \frac{π}{3}$

Show Answer

Solution

Since, $n = 10$ is even, so this expansion has only one middle term i.e., 6th term.

$\begin{aligned} ∴ T_{6} = T_{5 + 1} =^{10} C_{5} \frac{1}{x}^{10 - 5} (x \sin x)^{5} \\ \Rightarrow \frac{63}{8} =^{10} C_{5} x^{- 5} x^{5} \sin^{5} x \\ \Rightarrow \frac{63}{8} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 5!} \sin^{5} x \\ \Rightarrow \frac{63}{8} = 2 \cdot 9 \cdot 2 \cdot 7 \cdot \sin^{5} x \\ \Rightarrow \sin^{5} x = \frac{1}{32} \\ \Rightarrow \sin^{5} x = {\frac{1}{2}}^{5} \\ \Rightarrow \sin x = \frac{1}{2} \\ ∴ x = n π + (- 1)^{n} π / 6 \end{aligned}$

Fillers

25. The largest coefficient in the expansion of $(1 + x)^{30}$ is ……

Thinking Process

In the expansion of $(1 + x)^{n}$ , the largest coefficient is $^{n} C_{n / 2}$ (when $n$ is even).

Show Answer

Solution

Largest coefficient in the expansion of $(1 + x)^{30} =^{30} C_{30 / 2} =^{30} C_{15}$

26. The number of terms in the expansion of $(x + y + z)^{n}$ ……

Show Answer

Solution

Given expansion is $(x + y + z)^{n} = [x + (y + z)]^{n}$ .

$[x + (y + z)]^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} (y + z)$

$+^{n} C_{2} x^{n - 2} (y + z)^{2} + \dots +^{n} C_{n} (y + z)^{n}$

$∴$ Number of terms $= 1 + 2 + 3 + \dots + n + (n + 1)$

$= \frac{(n + 1) (n + 2)}{2}$

27. In the expansion of $x^{2} - {\frac{1}{x^{2}}}^{16}$ , the value of constant term is ……

Show Answer

Solution

Let constant be $T_{r + 1}$ .

$∴ T_{r + 1} =^{16} C_{r} (x^{2})^{16 - r} - \frac{1}{x^{2}}$

$=^{16} C_{r} x^{32 - 2 r} (- 1)^{r} x^{- 2 r}$ $=^{16} C_{r} x^{32 - 4 r} (- 1)^{r}$ $For constant term, 32 - 4 r = 0 \Rightarrow r = 8$ $∴ T_{8 + 1} =^{16} C_{8}$

28. If the seventh term from the beginning and the end in the expansion of $\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}}$ are equal, then $n$ equals to ……

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Solution

Given expansions is $\sqrt[3]{2} + {\frac{1}{\sqrt[3]{3}}}^{n}$

$∴ T_{7} = T_{6 + 1} =^{n} C_{6} (\sqrt[3]{2})^{n - 6} {\frac{1}{\sqrt[3]{3}}}^{6}$

Since, $T_{7}$ from end is same as the $T_{7}$ from beginning of $\frac{1}{\sqrt[3]{3}} + \sqrt[3]{2}^{n}$

Then,

$T_{7} =^{n} C_{6} {\frac{1}{\sqrt[3]{3}}}^{n - 6} (\sqrt[3]{2})^{6}$

Given that,

$^{n} C_{6} (2)^{\frac{n - 6}{3}} (3)^{- 6 / 3} =^{n} C_{6} (3)^{- \frac{(n - 6)}{3}} 2^{6 / 3}$

$\Rightarrow (2) \frac{n - 12}{3} = {\frac{1}{3^{1 / 3}}}^{n - 12}$

which is true, when $\frac{n - 12}{3} = 0$ .

$\Rightarrow$

$n - 12 = 0 \Rightarrow n = 12$

29. The coefficient of $a^{- 6} b^{4}$ in the expansion of $\frac{1}{a} - \frac{2 b}{3}$ is ……

Thinking Process

In the expansion of $(x - a)^{n}, T_{r + 1} =^{n} C_{r} x^{n - r} (- a)^{r}$

Show Answer

Solution

Given expansion is $\frac{1}{a} - {\frac{2 b}{3}}^{10}$ .

Let $T_{r + 1}$ has the coefficient of $a^{- 6} b^{4}$ .

$∴ T_{r + 1} =^{10} C_{r} {\frac{1}{a}}^{10 - r} - {\frac{2 b}{3}}^{r}$

For coefficient of $a^{- 6} b^{4}, 10 - r = 6 \Rightarrow r = 4$

Coefficient of $a^{- 6} b^{4} =^{10} C_{4} (- 2 / 3)^{4}$

$∴ = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6!}{6! \cdot 4 \cdot 3 \cdot 2 \cdot 1} \cdot \frac{2^{4}}{3^{4}} = \frac{1120}{27}$

30. Middle term in the expansion of $(a^{3} + b a)^{28}$ is ……

Show Answer

Solution

Given expansion is $(a^{3} + b a)^{28}$ .

$∵ n = 28$

[even]

$∴$ Middle term $= \frac{28}{2} + 1$ th term $= 15$ th term

$∴ T_{15} = T_{14 + 1}$

$=^{28} C_{14} (a^{3})^{28 - 14} (b a)^{14}$

$=^{28} C_{14} a^{42} b^{14} a^{14}$

$=^{28} C_{14} a^{56} b^{14}$

31. The ratio of the coefficients of $x^{p}$ and $x^{q}$ in the expansion of $(1 + x)^{p + q}$ is ……

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Solution

Given expansion is $(1 + x)^{p + q}$ .

$∴$ Coefficient of $x^{p} =^{p + q} C_{P}$

and coefficient of $x^{q} =^{p + q} C_{q}$

$∴ \frac{^{p + q} C_{p}}{^{p + q} C_{q}} = \frac{^{p + q} C_{p}}{^{p + q} C_{p}} = 1 : 1$

32. The position of the term independent of $x$ in the expansion of $\sqrt{\frac{x}{3}} + \frac{3}{2 x^{2}}$ is ……

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Solution

Given expansion is $\sqrt{\frac{x}{3}} + {\frac{3}{2 x^{2}}}^{10}$ .

Let the constant term be $T_{r + 1}$ .

Then,

$\begin{aligned} T_{r + 1} & =^{10} C_{r} {\sqrt{\frac{x}{3}}}^{10 - r} {\frac{3}{2 x^{2}}}^{r} \\ =^{10} C_{r} \cdot x^{\frac{10 - r}{2}} \cdot 3^{\frac{- 10 + r}{2}} \cdot 3^{r} \cdot 2^{- r} \cdot x^{- 2 r} \\ =^{10} C_{r} x^{\frac{10 - 5 r}{2}} 3^{\frac{- 10 + 3 r}{2}} 2^{- r} \end{aligned}$

For constant term, $10 - 5 r = 0 \Rightarrow r = 2$

Hence, third term is independent of $x$ .

33. If $25^{15}$ is divided by 13 , then the remainder is ……

Show Answer

Solution

Let

$\begin{aligned} 25^{15} & = (26 - 1)^{15} \\ =^{15} C_{0} 26^{15} -^{15} C_{1} 26^{14} + \dots -^{15} C_{15} \\ =^{15} C_{0} 26^{15} -^{15} C_{1} 26^{14} + \dots - 1 - 13 + 13 \\ =^{15} C_{0} 26^{15} -^{15} C_{1} 26^{14} + \dots - 13 + 12 \end{aligned}$

It is clear that, when $25^{15}$ is divided by 13 , then remainder will be 12 .

True/False

34. The sum of the series $\sum_{r = 0}^{10}^{20} C_{r}$ is $2^{19} + \frac{^{20} C_{10}}{2}$ .

Show Answer

Solution

False

Given series

$\begin{aligned} = \sum_{r = 0}^{10}^{20} C_{r} =^{20} C_{0} +^{20} C_{1} +^{20} C_{2} + \dots +^{20} C_{10} \\ =^{20} C_{0} +^{20} C_{1} + \dots +^{20} C_{10} +^{20} C_{11} + \dots^{20} C_{20} - (^{20} C_{11} + \dots +^{20} C_{20}) \\ = 2^{20} - (^{20} C_{11} + \dots +^{20} C_{20}) \end{aligned}$

Hence, the given statement is false.

35. The expression $7^{9} + 9^{7}$ is divisible by 64 .

Show Answer

Solution

True

Given expression $= 7^{9} + 9^{7} = (1 + 8)^{7} - (1 - 8)^{9}$

$\begin{aligned} = (^{7} C_{0} +^{7} C_{1} 8 +^{7} C_{2} 8^{2} + \dots +^{7} C_{7} 8^{7}) - (^{9} C_{0} -^{9} C_{1} 8 +^{9} C_{2} 8^{2} \dots -^{9} C_{9} 8^{9}) \\ = (1 + 7 \times 8 + 21 \times 8^{2} + \dots) - (1 - 9 \times 8 + 36 \times 8^{2} + \dots - 8^{9}) \\ = (7 \times 8 + 9 \times 8) + (21 \times 8^{2} - 36 \times 8^{2}) + \dots \\ = 2 \times 64 + (21 - 36) 64 + \dots \end{aligned}$

which is divisible by 64

Hence, the statement is true.

36. The number of terms in the expansion of $[(2 x + y^{3})^{4}]^{7}$ is 8 .

Show Answer

Solution

False

Given expansion is $[(2 x + y^{3})^{4}]^{7} = (2 x + y^{3})^{28}$ .

Since, this expansion has 29 terms.

So, the given statement is false.

37. The sum of coefficients of the two middle terms in the expansion of $(1 + x)^{2 n - 1}$ is equal to $^{2 n - 1} C_{n}$ .

Show Answer

Solution

False

Here, the Binomial expansion is $(1 + x)^{2 n - 1}$ .

Since, this expansion has two middle term i.e., $\frac{2 n - 1 + 1}{2}$ th term and $\frac{2 n - 1 + 1}{2} + 1$ th term i.e., $n$ th term and $(n + 1)$ th term.

$\begin{aligned} ∴ Coefficient of n th term & =^{2 n - 1} C_{n - 1} \\ Coefficient of (n + 1) th term & =^{2 n - 1} C_{n} \\ Sum of coefficients & =^{2 n - 1} C_{n - 1} +^{2 n - 1} C_{n} \\ =^{2 n - 1 + 1} C_{n} =^{2 n} C_{n} [∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}] \end{aligned}$

38. The last two digits of the numbers $3^{400}$ are 01.

Show Answer

Solution

True

Given that, $3^{400} = 9^{200} = (10 - 1)^{200}$

$\begin{matrix} \Rightarrow & (10 - 1)^{200} =^{200} C_{0} 10^{200} -^{200} C_{1} 10^{199} + \dots -^{200} C_{199} 10^{1} +^{200} C_{200} 1^{200} \\ \Rightarrow & (10 - 1)^{200} = 10^{200} - 200 \times 10^{199} + \dots - 10 \times 200 + 1 \end{matrix}$

So, it is clear that the last two digits are 01 .

39. If the expansion of $x - {\frac{1}{x^{2}}}^{2 n}$ contains a term independent of $x$ , then $n$ is a multiple of 2.

Show Answer

Solution

False

Given Binomial expansion is $x - {\frac{1}{x^{2}}}^{2 n}$ .

Let $T_{r + 1}$ term is independent of $x$ .

Then,

$\begin{aligned} T_{r + 1} & =^{2 n} C_{r} (x)^{2 n - r} - {\frac{1}{x^{2}}}^{r} \\ =^{2 n} C_{r} x^{2 n - r} (- 1)^{r} x^{- 2 r} =^{2 n} C_{r} x^{2 n - 3 r} (- 1)^{r} \end{aligned}$

For independent of $x$ ,

$∴ \begin{aligned} 2 n - 3 r & = 0 \\ r & = \frac{2 n}{3}, \end{aligned}$

which is not a integer.

So, the given expansion is not possible.

40. The number of terms in the expansion of $(a + b)^{n}$ , where $n \in N$ , is one less than the power $n$ .

Show Answer

Solution

False

We know that, the number of terms in the expansion of $(a + b)^{n}$ , where $n \in N$ , is one more than the power $n$ .

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