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Question

Prove that: x2ny2n is divisible by x+y.

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Solution

Let P(n):x2ny2n=(x+y)×d where dϵN

For n=1

LHS =x2×1y2×1
=x2y2
(x+y)(xy)
=RHS

P(n) is true for n=1

Assume P(k) is true.

x2ky2k=(x+y)×m where mϵN

We will prove that P(k+1) is true.

LHS =x2×(k+1)y2×(k+1)
=x2k+2y2k+2
=x2kx2y2ky2
=(x+y)[mx2+y2k(xy)]
=(x+y)×r

where, r=[mx2+y2k(xy)]

P(k+1) is true whenever P(k) is true.

By the principle of mathematical induction, P(n) is true for n, where n is a natual number.

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