Question

Sum of n terms of the series √2+√8+√18+√32.. is
A. n(n+1)/2
B. 2n (n + 1
C. n(n+1)/√2
D. 1

Answer 1

The required "option C) $\dfrac{n(n +1)}{\sqrt{2}}$" is correct.

Step-by-step explanation:

The given sum of n terms of the series:

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

i.e., $\sqrt{2}$ + 2$\sqrt{2}$ + 3$\sqrt{2}$ + 4$\sqrt{2}$ + ........

The given series are in AP.

Here, first term (a) = $\sqrt{2}$ , common difference (d) = 2$\sqrt{2}$ - $\sqrt{2}$ = $\sqrt{2}$

To find, the sum of n terms of the series = ?

We know that,

The sum of n terms of the A.P.

$S_{n}$ = $\dfrac{n}{2} [2a+(n-1)d]$

∴ The sum of n terms of the given series

= $\dfrac{n}{2}$ [2($\sqrt{2}$) + (n - 1)$\sqrt{2}$ ]

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

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

∴ The sum of n terms of the given series = $\dfrac{n(n +1)}{\sqrt{2}}$

Thus, the required "option C) $\dfrac{n(n +1)}{\sqrt{2}}$" is correct.