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

(d) $\frac{n+1}{n}$

Given:

Sum of the even natural numbers = k ×">× Sum of the odd natural numbers

$\frac{n}{2}\{2 a+(n-1) d\}=k \times \frac{n}{2}\{2 a+(n-1) d\}$

$\Rightarrow\{2 \times 2+(n-1) 2\}=k \times\{2 \times 1+(n-1) 2\}$

$\Rightarrow \frac{4+(n-1) 2}{2+(n-1) 2}=k$

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