Q63 / 200 · Manipal (MET) 2014 · Mathematics
Middle term in the expansion of $\left(x^2+\frac{1}{x^2}+2\right)^n$ is
- A$\frac{n!}{(n!)^2}$
- Correct answerB$\frac{(2 n!)}{(n!)^2}$
- C$\frac{(2 n-1)!}{n!}$
- D$\frac{(2 n)!}{n!}$
Solution
The given expression is
$\begin{aligned}
\left(x^2+\frac{1}{x^2}+2\right)^n & =\left[\left(x+\frac{1}{x}\right)^2\right]^n \\
& =\left(x+\frac{1}{x}\right)^{2 n}
\end{aligned}$
The binomial contains $(2 n+1)$ terms in its expansion.
$\therefore$ The middle term is $T_{n+1}$.
$\begin{aligned}
T_{n+1} & ={ }^{2 n} C_n(x)^{2 n-n}\left(\frac{1}{x}\right)^n \\
& ={ }^{2 n} C_n \\
& =\frac{(2 n)!}{(2 n-n)!n!}=\frac{(2 n)!}{(n!)^2}
\end{aligned}$
Hence, the correct answer is option: 'B'.