Question

Prove that $x^{2n-1}+y^{2n-1}$ is divisible by $x+y$ for all $n\in N$.

Answer 1

We shall prove the given statement by the principle of mathematical induction.
Let the given statement be P(n) : $x^{2n-14}+y^{2n-1}$ is divisible by $x+y$

Step I: For $n=1$

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

$=x+y$ is divisible by itself $x+y$

$\Rightarrow$ $P\left(n\right)$ is true for $n=1$

StepII: Assume $P\left(n\right)$ is true for $n=k$,

i.e., $x^{2k-1}+y^{2k-1}$ is divisible by $(x+y)$

$x^{2k-1}+y^{2k+1}=(x+y)\lambda$
$\Rightarrow x^{2k-1}=(x+y)\lambda -y^{2k+1}$ .......(1)

Step III: We shall prove the given statement true for $n=k+1$
i.e., It is sufficient to prove that
$x^{2k+1}+y^{2k+1}=(x+y)\mu$ for some $\mu$

Now, Consider the R.H.S of above equation
$=x^{2k+1}+y^{2k+1}$
$=x^{2k-1}.x^2+y^{2k-1}.y^2$
$=\left[\right[(x+y)\lambda -y^{2k-1}].x^2+y^{2k-1}.y^2]$ [ from equation(i)]

$=(x+y)\lambda x^2-y^{2k-1}.x^2+y^{2k-1}.y^2$

$=(x+y)\lambda x^2-y^{2k-1}(x^2-y^2)$

$=(x+y)\lambda x^2-y^{2k-1}(x+y)(x-y)$

$=(x+y)[\lambda x^2-y^{2k-1}(x-y\left)\right]$

$=(x+y)\mu$ where $\mu =\lambda x^2-y^{2k-1}(x-y)]$

Thus, $x^{2k+1}+y^{2k+1}$ is divisible by $x+y$

$\Rightarrow$ $P\left(n\right)$ is true for $n=k+1$

Hence, $P\left(n\right)$ is true for all $n\in N$ by principle of mathematical induction.