# Find the middle terms(s) in the expansion of:

Question:

Find the middle terms(s) in the expansion of:

(i) $\left(x-\frac{1}{x}\right)^{10}$

(ii) $\left(1-2 x+x^{2}\right)^{n}$

(iii) $\left(1+3 x+3 x^{2}+x^{3}\right)^{2 n}$

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

(v) $\left(x-\frac{1}{x}\right)^{2 n+1}$

(vi) $\left(\frac{x}{3}+9 y\right)^{10}$

(vii) $\left(3-\frac{x^{3}}{6}\right)^{7}$

(viii) $\left(2 a x-\frac{b}{x^{2}}\right)^{12}$

(ix) $\left(\frac{p}{x}+\frac{x}{p}\right)^{9}$

(X) $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$

Solution:

(i) $\left(x-\frac{1}{x}\right)^{10}$

Here, $n$ is an even number.

$\therefore$ Middle term $=\left(\frac{10}{2}+1\right)$ th $=6$ th term

Now, we have,

$T_{6}=T_{5+1}$

$={ }^{10} C_{5} x^{10-5}\left(\frac{-1}{x}\right)^{5}$

$=-\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2}$

$=-252$

(ii) $\left(1-2 x+x^{2}\right)^{n}$

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

$n$ is an even number.

$\therefore$ Middle term $=\left(\frac{2 n}{2}+1\right)$ th $=(n+1)$ th term

Now, we have,

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

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

(iii) $\left(1+3 x+3 x^{2}+x^{3}\right)^{2 n}$

$=(1+x)^{6 n}$

Here, $n$ is an even number.

$\therefore$ Middle term $=\left(\frac{6 n}{2}+1\right)$ th $=(3 n+1)$ th term

Now, we have,

$T_{3 n+1}$

$={ }^{6 n} C_{3 n} x^{3 n}$

$=\frac{(6 n) !}{(3 n !)^{2}} x^{3 n}$

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

Here, $n$ is an odd number.

Therefore, the middle terms are $\left(\frac{n+1}{2}\right)$ th and $\left(\frac{n+1}{2}+1\right)$ th, i. e. 5 th and 6 th terms.

Now, we have

$T_{5}=T_{4+1}$

$={ }^{9} C_{4}(2 x)^{9-4}\left(\frac{-x^{2}}{4}\right)^{4}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 2^{5} \frac{1}{4^{4}} x^{5+8}$

$=\frac{63}{4} x^{13}$

And,

$T_{6}=T_{5+1}$

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

$=-\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 2^{4} \frac{1}{4^{5}} x^{4+10}$

$=-\frac{63}{32} x^{14}$

(v) $\left(x-\frac{1}{x}\right)^{2 n+1}$

Here, $(2 n+1)$ is an odd number.

Therefore, the middle terms are $\left(\frac{2 n+1+1}{2}\right)$ th and $\left(\frac{2 n+1+1}{2}+1\right)$ th i. e. $(n+1)$ th and $(n+2)$ th terms.

Now, we have :

$T_{n+1}$

$={ }^{2 n+1} C_{n} x^{2 n+1-n} \times \frac{(-1)^{n}}{x^{n}}$

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

And,

$T_{n+2}=T_{n+1+1}$

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

$=(-1)^{n+1}{ }^{2 n+1} C_{n+1} \times \frac{1}{x}$

(vi) $\left(\frac{x}{3}+9 y\right)^{10}$

Here, $n$ is an even number.

Therefore, the middle term is $\left(\frac{10}{2}+1\right)$ th, i.e., 6 th term.

Now, we have

$T_{6}=T_{5+1}$

$={ }^{10} C_{5}\left(\frac{x}{3}\right)^{10-5}(9 y)^{5}$

$=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} \times \frac{1}{3^{5}} \times 9^{5} \times x^{5} y^{5}$

$=61236 x^{5} y^{5}$

(vii) $\left(3-\frac{x^{3}}{6}\right)^{7}$

Here, $n$ is an odd number.

Therefore, the middle terms are $\left(\frac{7+1}{2}\right)$ th and $\left(\frac{7+1}{2}+1\right)$ th, i.e., 4 th and 5 th terms.

Now, we have

$T_{4}=T_{3+1}$

$={ }^{7} C_{3}(3)^{7-3}\left(\frac{-x^{3}}{6}\right)^{3}$

$=-\frac{105}{8} x^{9}$

And,

$T_{5}=T_{4+1}$

$={ }^{9} C_{4}(3)^{9-4}\left(\frac{-x^{3}}{6}\right)^{4}$

$=\frac{7 \times 6 \times 5}{3 \times 2} \times 3^{5} \times \frac{1}{6^{4}} x^{12}$

$=\frac{35}{48} x^{12}$

(viii) $\left(2 a x-\frac{b}{x^{2}}\right)^{12}$

Here, $n$ is an even number.

$\therefore$ Middle term $=\left(\frac{12}{2}+1\right)^{\text {th }}=7^{\text {th }}$ term

Now, we have

$T_{7}=T_{6+1}$

$={ }^{12} C_{6}(2 a x)^{12-6}\left(\frac{-b}{x^{2}}\right)^{6}$

$=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \times\left(\frac{2 a b}{x}\right)^{6}$

$=\frac{59136 a^{6} b^{6}}{x^{6}}$

(ix) $\left(\frac{p}{x}+\frac{x}{p}\right)^{9}$

Here, $n$ is an odd number.

Therefore, the middle terms are $\left(\frac{9+1}{2}\right)^{\text {th }}$ and $\left(\frac{9+1}{2}+1\right)^{\text {th }}$, i.e., $5^{\text {th }}$ and $6^{\text {th }}$ terms.

Now, we have

$T_{5}=T_{4+1}$

$={ }^{9} C_{4}\left(\frac{p}{x}\right)^{9-4}\left(\frac{x}{p}\right)^{4}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times\left(\frac{p}{x}\right)$

$=\frac{126 p}{x}$

And,

$T_{6}=T_{5+1}$

$={ }^{9} C_{5}\left(\frac{p}{x}\right)^{9-5}\left(\frac{x}{p}\right)^{5}$

$=\frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times\left(\frac{x}{p}\right)$

$=\frac{126 x}{p}$

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

Here, $n$ is an even number.

$\therefore$ Middle term $=\left(\frac{10}{2}+1\right)^{\text {th }}=6^{\text {th }}$ term

Now, we have

$T_{6}=T_{5+1}$

$={ }^{10} C_{5}\left(\frac{x}{a}\right)^{10-5}\left(\frac{-a}{x}\right)^{5}$

$=-\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$

$=-252$