A) \[\frac{n}{{{[(n/2)!]}^{2}}}\]
B) \[-\frac{(2n!)}{{{[(n/2)!]}^{2}}}\]
C) \[\frac{1\cdot 3\cdot 5...(2n+1)}{n!}{{2}^{n}}\]
D) \[\frac{(2n)!}{{{(n!)}^{2}}}\]
Correct Answer: D
Solution :
\[\because \]\[{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}+2 \right)}^{n}}={{\left[ {{\left( x+\frac{1}{x} \right)}^{2}} \right]}^{n}}={{\left( x+\frac{1}{x} \right)}^{2n}}\] The number of terms in the expansion of\[{{\left( x+\frac{1}{x} \right)}^{2n}}\]is\[2n+1\]which is odd. Therefore,\[(n+1)\text{th}\]term will be middle term. \[\therefore \] \[{{T}_{n+1}}{{=}^{2n}}{{C}_{n}}{{(x)}^{2n-n}}{{\left( \frac{1}{x} \right)}^{n}}\] \[{{=}^{2n}}{{C}_{n}}{{x}^{n}}\cdot \frac{1}{{{x}^{n}}}\] \[{{=}^{2n}}{{C}_{n}}=\frac{(2n)!}{n!n!}=\frac{(2n)!}{{{(n!)}^{2}}}\]
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