UPSC NDA II 2015 Mathematics Paper

Concept:The given series is an arithmetic progression (AP) after simplifying each term under the square root.Explanation:Write each term in simplest radical form:$\sqrt{2}$, $\sqrt{8} = 2\sqrt{2}$, $\sqrt{18} = 3\sqrt{2}$, $\sqrt{32} = 4\sqrt{2}$, …The series becomes: $\sqrt{2} + 2\sqrt{2} + 3\sqrt{2} + 4\sqrt{2} + \ldots$This is an AP with first term $a = \sqrt{2}$ and common difference $d = \sqrt{2}$.Sum of $n$ terms of an AP is $S_n = \frac{n}{2}[2a + (n-1)d]$.Substitute $a$ and $d$:$S_n = \frac{n}{2}[2\sqrt{2} + (n-1)\sqrt{2}]$Factor $\sqrt{2}$ inside brackets:$S_n = \frac{n}{2} \cdot \sqrt{2}[2 + (n-1)] = \frac{n}{2} \cdot \sqrt{2}(n+1)$Simplify:$S_n = \frac{n(n+1)\sqrt{2}}{2} = \frac{n(n+1)}{\sqrt{2}}$ after rationalizing? Wait check: $\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$, so indeed $S_n = \frac{n(n+1)}{\sqrt{2}}$.Answer:$\displaystyle \frac{{\rm{n}}({\rm{n}}+1)}{\sqrt{2}}$ (Option C).