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Mathematics Question on binomial expansion formula

The 7th7^{th} term in the expansion of (x2+1x2+2)n\left(x^{2}+\frac{1}{x^{2}}+2\right)^{n} is

A

n![n/5!]2x2\frac{n!}{\left[n/5!\right]^{2}} x^{2}

B

nC6 x6^nC_6\,x^6

C

1.3.5...(2n+1)n!2n\frac{1.3.5...\left(2n+1\right)}{n!} 2^{n}

D

None of these

Answer

None of these

Explanation

Solution

We have, (x2+1x2+2)n=(x+1x)2n\left(x^{2}+\frac{1}{x^{2}}+2\right)^{n} = \left(x+\frac{1}{x}\right)^{2n} ∴T7=T6+1= 2nC6(x)2n−6(1x)6\therefore T_{7} = T_{6+1} = \,^{2n}C_{6}\left(x\right)^{2n-6} \left(\frac{1}{x}\right)^{6} = 2nC6 x2n−12= \,^{2n}C_{6}\,x^{2n-12}