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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
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Solution(By Apex Team)

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

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