Question

If the sum of first n even natural numbers is equal to k times the sum of the 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 1

The correct option is D

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

Given:

Sum of the even natural numbers = k $\times$ Sum of the odd natural numbers.

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

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

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

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