Question

The sum of its n- terms of the series √2 + √8 +√ 18 +√ 32 +…… is

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

Sum of first n terms of an arithmetic progression

$= \displaystyle \: \frac{n}{2} [2a + (n - 1)d ]$

Where First term = a

Common Difference = d

CALCULATION

Here

$\sqrt{8} = \sqrt{2 \times 2 \times 2} = 2 \sqrt{2}$

$\sqrt{18} = \sqrt{3 \times 3 \times 2} = 3 \sqrt{2}$

$\sqrt{32} = \sqrt{2 \times 2 \times 2 \times 2 \times 2} = 4 \sqrt{2}$

So we have to sum n terms of the series

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

First term = a =

$\sqrt{2}$

Common Difference = d =

$2 \sqrt{2} - \sqrt{2} = \sqrt{2}$

So the required sum is

$= \displaystyle \: \frac{n}{2} [2a + (n - 1)d ]$

$= \displaystyle \: \frac{n}{2} [2 \sqrt{2} + (n - 1) \times \sqrt{2} ]$

$= \displaystyle \: \frac{n}{2} [ \sqrt{2} + n \sqrt{2} ]$

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

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