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Sum of n terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+$…. is?

A. $\Large\frac{n(n+1)}{2}$ B. $2 n(n+1)$ C. $\Large\frac{n(n+1)}{\sqrt{2}}$ D. 1 Answer: Option C
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Solution(By Apex Team)

The series is given $\begin{array}{l} \sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ldots \ldots \\ \Rightarrow \sqrt{2}+2 \sqrt{2}+3 \sqrt{2}+4 \sqrt{2}+\ldots \ldots \end{array}$ Here a = $\sqrt{2}$ and $\mathrm{d}=2 \sqrt{2}-\sqrt{2}=\sqrt{2}$ $\begin{aligned}\therefore S_n&=\frac{n}{2}[2a+(n-1)d]\\ &=\frac{n}{2}[2\sqrt{2}+(n-1)\sqrt{2}]\\ &=\frac{n}{2}[2\sqrt{2}+\sqrt{2}n-\sqrt{2}]\\ &=\frac{n}{2}(\sqrt{2}n+\sqrt{2})\\ &=\frac{n\sqrt{2}}{2}(n+1)\\ &=\frac{n(n+1)}{\sqrt{2}}\end{aligned}$

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