### 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. $\frac{n+1}{2 n}$ D. $\frac{n+1}{n}$ Answer: Option D

### Solution(By Apex Team)

Sum of n even natural number = n(n+1) and sum of n odd natural numbers = $n^{2}$ \begin{aligned}\therefore n(n+1)&=kn^2\\ \Rightarrow k&=\frac{n(n+1)}{n^2}\\ &=\frac{n+1}{n}\end{aligned}

A. 22
B. 25
C. 23
D. 24

A. 5
B. 6
C. 4
D. 3

A. -45
B. -55
C. -50
D. 0