The coefficient of xn in the expansion of (1+x)(1−x)n is
A
(−1)n−1n
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B
(−1)n(1−n)
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C
(−1)n−1(n−2)2
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D
(n−1)
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Solution
The correct option is C(−1)n(1−n) We have Coefficient of xn in (1+x)(1−x)n = Coefficient of xn in (1−x)n+ Coefficient of xn−1 in (1−x)n =(−1)nnCn+(−1)n−1nCn−1=(−1)n(1−n)