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

The correct option is C $\frac{n+1}{2n}$

$\text{solution:}{n}^{2}k=\frac{n(n+1)}{2}=\text{sum of first}n\text{'}$

$\text{Sum of first}n\text{add}=1+3+5+\dots .\text{nth numbers}$

$=\frac{n}{2}(2x1+(n-1\left)2\right)$

$=\frac{n}{2}\left(2n\right)=n^2$

$\Rightarrow k=\frac{n^2+n}{2n^2}=\frac{n+1}{2n}$