Question

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

• A. x + y
• B. x - y
• C. $x^2+y^2$
• D. $x^2+xy$

Answer 1

The correct option is A x + y

Given expression is $x^{2n-1}+y^{2n-1}$, and also $n$ is a natural number,

Consider the statement $S\left(p\right)$ that ${ x }^{ 2n-1 }+{ y }^{ 2n-1 }$isdivisibleby$p$,

$1)$ let $n=1$,

$⟹x+y$, this is divisible by $x+y$,

$\LongrightarrowS\left( 1 \right) $ is true

$2)$ Let us assume that $S\left(n\right)$ is true,

$⟹$ if we replace $n=n+1$,

The statement becomes,

$x^{2n+1}+y^{2n+1}$ is also divisible by $x+y$,

$\therefore$ by the principle of mathematical induction $x^{2n-1}+y^{2n-1}$ is divisible by $x+y$