Question:
Sum of $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ldots .$ is
(a) $\frac{n(n+1)}{2}$
(b) $2 n(n+1)$
(c) $\frac{n(n+1)}{\sqrt{2}}$
(d) 1
Solution:
(c) $\frac{n(n+1)}{\sqrt{2}}$
Let $T_{n}$ be the $n$th term of the given series.
Thus, we have:
$T_{n}=\sqrt{2 \times n^{2}}=n \sqrt{2}$
Now, let $S_{n}$ be the sum of $n$ terms of the given series.
Thus, we have:
$S_{n}=\sqrt{2} \sum_{k=1}^{n}(k)$
$\Rightarrow S_{n}=\sqrt{2}\left[\frac{n(n+1)}{2}\right]$
$\Rightarrow S_{n}=\frac{n(n+1)}{\sqrt{2}}$
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