Sum of n terms of the series 2 + 8 + 18 + 32 + ....... is

Answer: Option C

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