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(D) Let $S_1$ be the sum of the first $n$ even natural numbers.$S_1 = 2 + 4 + 6 + \dots + 2n$This is an arithmetic progression with $a = 2$,$d = 2$,an

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

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(D) Let $S_1$ be the sum of the first $n$ even natural numbers.$S_1 = 2 + 4 + 6 + \dots + 2n$This is an arithmetic progression with $a = 2$,$d = 2$,and $n$ terms.$S_1 = \frac{n}{2} [2(2) + (n-1)2] = \frac{n}{2} [4 + 2n - 2] = \frac{n}{2} [2n + 2] = n(n+1)$.Let $S_2$ be the sum of the first $n$ odd natural numbers.$S_2 = 1 + 3 + 5 + \dots + (2n-1)$This is an arithmetic progression with $a = 1$,$d = 2$,and $n$ terms.$S_2 = \frac{n}{2} [2(1) + (n-1)2] = \frac{n}{2} [2 + 2n - 2] = \frac{n}{2} [2n] = n^2$.Given that $S_1 = k 	imes S_2$,we have:$n(n+1) = k 	imes n^2$$k = \frac{n(n+1)}{n^2} = \frac{n+1}{n}$.

If the sum of the first $n$ even natural numbers is $k$ times the sum of the first $n$ odd natural numbers,then $k = ........$

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