Question

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

• 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 $\frac{n+1}{n}$

We know,

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$ ......... Sum of n terms

$a=$ First term

$d=$ Common difference

$n=$ number of terms

First n even natural numbers are $:2,4,6,\mathrm{8......}$

It forms an AP with first term $=2$ and common difference $=2$

Sum $=\frac{n}{2}\left(2\right(2)+(n-1\left)2\right)=\frac{2n(n+1)}{2}$

First n odd natural numbers are $:1,3,5,7,\mathrm{9........}$

It forms an AP with first term $=1$ and common difference $=2$

Sum $=\frac{n}{2}\left(2\right(1)+(n-1\left)2\right)=\frac{2n^2}{2}$

According to the question,

$K\frac{2n^2}{2}=2\cdot \frac{n^2+n}{2}$

$\Rightarrow n^2(K-1)=n$

$\therefore K=\frac{n+1}{n}$