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Prove the following by using the principle of mathematical induction for all $n \in N$ \[\] $ x^{2n}-y^{2n}$ is divisible by $x+y$.

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Let the given statement be $P(n)$, i.e.,
$P(n) : x^{2n}-y^{2n}$ is divisible by $x+y$
It can be observed that $P(n)$ is true for $n=1$
This is so because $x^{ 2 \times 1} - y^{2 \times 1} = x^2 - y^2= (x+y)(x-y)$ is divisible by $(x+y)$
Let $P(k)$ be true for some positive integer $k$, i.e.,
$x^{2k}-y^{2k}$ is divisible by $x+y$
$ \therefore x^{2k}-y^{2k}= m(x+y)$, where $m \in N$--------(1)
We shall now prove that $P(k+1)$ is true whenever $P(k) $ is true.
$x^{2(k+1)}- y^{2(k+1)}$
$ = x^{2k}.x^2-y^{2k}.y^2$
$ = x^2 (x^{2k}-y^{2k}+y^{2k})-y^{2k}.y^2$
$ = x^2 \{m(x+y) +y^{2k} \}-y^{2k}.y^2\qquad$ [ Using (1)]
$= m(x+y)x^2+y^{2k}.x^2-y^{2k}.y^2$
$= m(x+y)x^2+y^{2k}(x^2-y^2)$
$= m(x+y)x^2+y^{2k}(x+y)(x-y)$
$ = (x+y) \{mx^2+y^{2k}(x-y) \}$ which is a factor of $(x+y)$.
Thus, $P(k+1)$ is true whenever $P(k)$ is true.
Hence, by the principle of mathematical induction, statement $P(n)$ is true for all natural numbers i.e., $n$.
answered May 1, 2014 by thanvigandhi_1

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