# Sum of n terms of the series 2–√+8–√+18−−√+32−−√+..is

Question:

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

Solution:

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.

= 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).