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Question

Mathematics Question on Binomial theorem

The coefficient of xnx^n in expansion of (1+x)(1−x)n(1+ x)(1- x)^n is

A

(−1)n−1n(- 1)^{n-1} n

B

(−1)n(1−n)(- 1)^n (1 -n)

C

(−1)n−1(n−1)2(-1)^{n-1} (n-1)^2

D

(n−1)(n - 1 )

Answer

(−1)n(1−n)(- 1)^n (1 -n)

Explanation

Solution

Coeff. of xnx^{n} in (1+x)(1−x)n \left(1+x\right)\left(1-x\right)^{n} = coeff of xnx^n in (1+x)(1−nC1x+nC2x2−...+(−1)n× nCnxn)\left(1+x\right)\left(1 - {^{n}C_{1}}x + {^{n}C_{2}}x^{2} - ...+\left(-1\right)^{n} \times \, ^nC_{n} x^{n}\right) =(−1)n× nCn+(−1)n−1 nCn−1=(−1)n+(−1)n−1.n= \left(-1\right)^{n} \times \,{^{n}C_{n} }+ \left(-1\right)^{n-1}\, {^{n}C_{n-1}} = \left(-1\right)^{n} + \left(-1\right)^{n-1} .n =(−1)n(1−n) = \left(-1\right)^{n}\left(1-n\right)