# Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ :Q: 21        $x^2^n-y^2^n$ is divisible by  $x+y.$

Let the given statement be p(n) i.e.
$p(n):x^2^n-y^2^n$
For n = 1  we have
$p(1):x^{2(1)}-y^{2(1)}= x^2-y^2=(x-y)(x+y)$   ,   which is divisible by   $(x+y)$ , hence true                $(using \ a^2-b^2=(a+b)(a-b))$

For  n = k  we have

$p(k):x^{2k}-y^{2k} \ \ \ \ \ \ \ \ \ \ -(i)$   ,        Let's assume that this is divisible by $(x+y)$   $=(x+y)m$

Now,
For  n = k + 1  we have
$p(k+1):x^{2(k+1)}-y^{2(k+1)}$                                                                                                                                                                                                                                                                                             $=x^{2k}.x^2-y^{2k}.y^2$

$=x^2(x^{2k}+y^{2k}-y^{2k})-y^{2k}.y^2$
$=x^2(x^{2k}-y^{2k})+x^2.y^{2k}-y^{2k}.y^2$
$=x^2(x+y)m+(x^2-y^2)y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ (i))$
$=x^2(x+y)m+((x-y)(x+y))y^{2k} \ \ \ \ \ \ \ \ \ \ \ \ (using \ a^2-b^2=(a+b)(a-b))$
$=(x+y)\left ( x^2.m+(x-y).y^{2k} \right )$
$=(x+y)l$                         where   $l = (x^2.m+(x-y).y^{2k})$     some natural number

Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is divisible by $(x+y)$  for all natural numbers n

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