For all positive integers n > 1, xxn−1−nan−1+an(n−1) is divisible by

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(x−a)2

x-a

2(x-a)

x+a

answer is B.

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Detailed Solution

LetP(n):xxn−1−nan−1+an(n−1), where n>1P(2):x(x−2a)+a2=(x−a)2P(3):xx2−3a2+2a3=x3−3a2x+2a3 P(3)=(x−a)x2+xa−2a2which are divisible by (x - a)

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