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Middle term in the expansion of $\left(x^{2}+\frac{1}{x^{2}}+2\right)^{n}$ is
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Verified Answer
The correct answer is:
$\frac{(2 n) !}{(n !)^{2}}$
The given expression is
$$
\left(x^{2}+\frac{1}{x^{2}}+2\right)^{n}=\left[\left(x+\frac{1}{x}\right)^{2}\right]^{n}=\left(x+\frac{1}{x}\right)^{2 n}
$$
The binomial contains $(2 n+1)$ terms in its expansion.
$\therefore$ The middle terms is $T_{n+1}$.
$$
T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(\frac{1}{x}\right)^{n}={ }^{2 n} C_{n}=\frac{(2 n) !}{(2 n-n) ! n !}=\frac{(2 n) !}{(n !)^{2}}
$$
$$
\left(x^{2}+\frac{1}{x^{2}}+2\right)^{n}=\left[\left(x+\frac{1}{x}\right)^{2}\right]^{n}=\left(x+\frac{1}{x}\right)^{2 n}
$$
The binomial contains $(2 n+1)$ terms in its expansion.
$\therefore$ The middle terms is $T_{n+1}$.
$$
T_{n+1}={ }^{2 n} C_{n}(x)^{2 n-n}\left(\frac{1}{x}\right)^{n}={ }^{2 n} C_{n}=\frac{(2 n) !}{(2 n-n) ! n !}=\frac{(2 n) !}{(n !)^{2}}
$$
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