If $x + y = a + b, x^{2} + y^{2} = a^{2} + b^{2}$ then prove that the mathematical induction that $x^{n} + y^{n} = a^{n} + b^{n}$ for all the natural number n .

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Solution

Clearly p(1) and p(2) hold. Assume p(n).
$p (n + 1) =$ $x^{n + 1} + y^{n + 1}$ = x. $x^{n}$ + y. $y^{n}$
$= x (a^{n} + b^{n} - y^{n}) + y . (a^{n} + b^{n} - x^{n})$
$= (a^{n} + b^{n})$ $(x + y) - x y$ $(x^{n - 1} + y^{n - 1})$
Now from given relations
$(x + y)^{2} - (x^{2} + y^{2}) = (a + b)^{2} - (a^{2} + b^{2})$
$∴$ $2 x y = 2 a b o r x y = a b$
Hence from (1)
$p (n + 1) =$ $(a^{n} + b^{n})$ $(a + b)$ - $a b (a^{n - 1} + b^{n - 1})$ by p(n - 1)
$= a^{n + 1} + b^{n + 1}$
Above shows that $p (n + 1)$ also holds good.

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