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 =
(a) $\frac{1}{n}$

(b) $\frac{n-1}{n}$

(c) $\frac{n+1}{2n}$

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

Answer 1

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

Given:
$2+4+...+2n=k\left[1+3+5+7+...+\left(2n-1\right)\right]$

$$\begin{gathered} \Rightarrow 2\times \frac{n\left(n+1\right)}{2}=k\left[\frac{n}{2}\left\{2\times 1+\left(n-1\right)\times 2\right\}\right] \\ \Rightarrow \frac{2n\left(n+1\right)}{2}=k\left[\frac{n}{2}\left(2+2n-2\right)\right] \\ \Rightarrow n\left(n+1\right)=k\left[\frac{n}{2}\left(2n\right)\right] \\ \Rightarrow n^2+n=k{n}^2 \\ \Rightarrow k=\frac{n+1}{n} \end{gathered}$$