Question

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

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n+1)}{\sqrt{2}}$
• C. $2n(n+1)$
• D. $1$

Answer 1

The correct option is B $\frac{n(n+1)}{\sqrt{2}}$

$\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+....$

$=\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+.....+n\sqrt{2}$

$=\sqrt{2}(1+2+3+4+.....+n)$

By using ${\sum }_{i=1}^{n}k=1+2+3+\dots +n=\frac{n(n+1)}{2}$

$=\sqrt{2}\left[\frac{n(n+1)}{2}\right]$

$=\frac{n(n+1)}{\sqrt{2}}$