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

A. $$\frac{1}{n}$$

B. $$\frac{{n - 1}}{n}$$

C. $$\frac{{n + 1}}{{2n}}$$

D. $$\frac{{n + 1}}{n}$$

Answer: Option D

Solution(By Examveda Team)

Sum of n even natural number = n(n+1)
and sum of n odd natural numbers = n²
$$\eqalign{ & \therefore n\left( {n + 1} \right) = k{n^2} \cr & \Rightarrow k = \frac{{n(n + 1)}}{{{n^2}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n + 1}}{n} \cr} $$