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Question

Mathematics Question on Binomial theorem

The middle term in the expansion of (1−1x)n(1−xn)\left(1-\frac{1}{x}\right)^{n} \left(1-x^{n}\right) in powers of x is

A

−2nCn−1-^{2n}C_{n-1}

B

−2nCn-^{2n}C_{n}

C

2nCn−1^{2n}C_{n-1}

D

2nCn^{2n}C_{n}

Answer

2nCn^{2n}C_{n}

Explanation

Solution

Given expansion can be re-written as (1−1x)n.(1−xn)=(−1)nx−n(1−x)2n\left(1-\frac{1}{x}\right)^{n} .\left(1-x^{n}\right) = \left(-1\right)^{n}x^{-n} \left(1-x\right)^{2n} Total number of terms will be 2n+12n + 1 which is odd (∵2n\because 2n is always even) ∴\therefore Middle term =2n+1+12=(n+1) th= \frac{2n+1+1}{2} = \left(n+1\right) \,th Now, Tr+1= nCr(1)rxn−rT_{r+1}=\,^{n}C_{r}\left(1\right)^{r} x^{n-r} So, 2nCn.x2n−nxn.(−1)n=2nCn.(−1)n\frac{^{2n}C_{n} . x^{2n-n}}{x^{n}.\left(-1\right)^{n}} = ^{2n}C_{n} .\left(-1\right)^{n} Middle term is an odd term. So, n+1n + 1 will be odd. So, n will be even. ∴\therefore Required answer is 2nCn.^{2n}C_{n}.