The coefficient of xn in expansion of (1+x)(1−x)n, is
A
(n−1)
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B
(−1)n(1−n)
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C
(−1)n−1(n−1)2
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D
(−1)n−1n
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Solution
The correct option is C(−1)n(1−n) The coefficient of xn in expansion of (1+x)(1−x)n = Coefficient of xn+ Coefficient of xn−1 =(−1)nn!n!0!+(−1)n−1n!1!(n−1)! =(−1)n[n!n!0!−n!1!(n−1)!] =(−1)n[n!n!0!−n!1!(n−1)!] =(−1)n(1−n)