Question

The sum of $24$ terms of the following series $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+...$ is

• A. $300$
• B. $200\sqrt{2}$
• C. $300\sqrt{2}$
• D. $250\sqrt{2}$

Answer 1

The correct option is C

$300\sqrt{2}$

Explanation for the correct option

Given series: $\sqrt{2}+\sqrt{8}+\sqrt{18}+\sqrt{32}+...$

$\Rightarrow$ $1\sqrt{2}+2\sqrt{2}+3\sqrt{2}+4\sqrt{2}+...$

$\Rightarrow$ $\sqrt{2}\left(1+2+3+4+...\right)$

Now, $\left(1+2+3+4+...\right)$ is forming an A.P. with first term $\mathrm{a}=1$, number of terms $\mathrm{n}=24$ and common difference $\mathrm{d}=1$, as it is a series of consecutive natural numbers upto $24$

Sum of $\mathrm{n}$ terms of an A.P. is ${\mathrm{S}}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}\right]$

Here, ${\mathrm{S}}_{24}=\frac{24}{2}\left[2\left(1\right)+\left(24-1\right)\left(1\right)\right]$

$\Rightarrow$ ${\mathrm{S}}_{24}=12\left[2+23\right]$

$\Rightarrow$ ${\mathrm{S}}_{24}=12\times 25$

$\Rightarrow$ ${\mathrm{S}}_{24}=300$

So, $\left(1+2+3+4+...\right)=300$

Thus, $\sqrt{2}\left(1+2+3+4+...\right)=300\sqrt{2}$

Hence, the correct option is option(C)