Question

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

Answer 1

Let $T_n$ be the nth terms 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 : $S_n=\sqrt{2}{\sum }_{k=1}^n\left(k\right)\Rightarrow S_n=\sqrt{2}\left[\frac{n(n+1)}{\sqrt{2}}\right]\Rightarrow S_n=\frac{n(n+1)}{\sqrt{2}}$