Sum of $n$ terms of the series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+\ldots$ is
(a) $\frac{n(n+1)}{2}$
(b) $2 n(n+1)$
(c) $\frac{n(n+1)}{\sqrt{2}}$
(d) 1
In the given problem, we need to find the sum of terms for a given arithmetic progression,
$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots$
So, here we use the following formula for the sum of n terms of an A.P.,
$S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Where; a = first term for the given A.P.
d = common difference of the given A.P.
n = number of terms
Here,
Common difference of the A.P. (d) =
$=\sqrt{8}-\sqrt{2}$
$=2 \sqrt{2}-\sqrt{2}$
$=\sqrt{2}$
Number of terms (n) = n
First term for the given A.P. (a) =
So, using the formula we get,
$S_{n}=\frac{n}{2}[2(\sqrt{2})+(n-1)(\sqrt{2})]$
$=\left(\frac{n}{2}\right)[2 \sqrt{2}+(\sqrt{2} n-\sqrt{2})]$
$=\left(\frac{n}{2}\right)(2 \sqrt{2}+\sqrt{2} n-\sqrt{2})$
$=\left(\frac{n}{2}\right)(\sqrt{2}+\sqrt{2} n)$
Now, taking $\sqrt{2}$ common from both the terms inside the bracket we get,
$=\left(\frac{n}{2}\right) \sqrt{2}(n+1)$
$=\frac{n(n+1)}{\sqrt{2}}$
Therefore, the sum of first $n$ terms for the given A.P. is $S_{n}=\frac{n(n+1)}{\sqrt{2}}$. So, the correct option is (c).