If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers,

According to the question,

$2+4+\ldots+2 n=k(1+3+5+7+\ldots+(2 n-1))$

$\Rightarrow 2 \times \frac{n(n+1)}{2}=k\left[\frac{n}{2}\{2 \times 1+(n-1) \times 2\}\right]$

$\Rightarrow \frac{2 n(n+1)}{2}=k\left[\frac{n}{2}(2+2 n-2)\right]$

$\Rightarrow n(n+1)=k\left[\frac{n}{2}(2 n)\right]$

$\Rightarrow n^{2}+n=k n^{2}$

$\Rightarrow k=\frac{n+1}{n}$