The coefficient of xn in expansion of (1 + x)(1 – x)n is
(a) (–1)n – 1 (n – 1)2 

(b) (–1)n (1 – n)

(c) (n – 1) 

(d) (–1)n – 1n.

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1 Answers

(b): (1 + x) (1 – x)n = (1 – x)n + x(1 – x)n
Therefore, Coefficient of xn is = (–1)n + (–1)n – 1nC1
= (–1)n[1 – n]

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