Question

If $n$ $\in$ N, then $x^{2n-1}+y^{2n-1}$ is divisible by

• A. $x+y$
• B. $x-y$
• C. $x^2+y^2$
• D. none of these

Answer 1

Answer: (A)

$x+y$

Le4 the given statement be $P\left(n\right)$

$P\left(n\right)=x^{2n-1}+y^{2n-1}$

We check $P\left(n\right)$ for $n=1$

$P\left(1\right)=x^{2-1}+y^{2-1}=x+y$

Thus $P\left(1\right)$ is divisible by $x+y$

Let us assume that $P\left(n\right)$ is divisible by $x+y$ when $n=k$

Now for $n=k+1$ we have

$P(k+1)=x^{2k+1}+y^{2k+1}$

$\Rightarrow P(k+1)=\left(x^2\right)\left(x^{2k-1}\right)+\left(y^2\right)\left(y^{2k-1}\right)$

$\Rightarrow P(k+1)=\left(x^2\right)(x^{2k-1}-y^2)(x^{2k-1}+y^2)\left(x^{2k-1}\right)+\left(y^2\right)\left(y^{2k-1}\right)$

$\Rightarrow P(k+1)=(x^2-y^2)\left(x^{2k-1}\right)+\left(y^2\right)(y^{2k-1}+x^{2k-1})$

$\Rightarrow P(k+1)=(x-y)(x+y)\left(x^{2k-1}\right)+\left(y^2\right)(y^{2k-1}+x^{2k-1})$

Now $(x-y)(x+y)\left(x^{2k-1}\right)$ and $\left(y^2\right)(y^{2k-1}+x^{2k-1})$ are divisible by $x+y$

so $x^{2k+1}+y^{2k+1}$ is also divisible by $x+y$

So $x^{2n-1}+y^{2n-1}$ is divisible by $x+y$ when $n=k+1$ if it is divisible by $x+y$ when $n=k$

so by principle of mathematical induction, $x^{2n-1}+y^{2n-1}$ is divisible by $x+y$