# The sum of 10 terms of the series

Question:

The sum of 10 terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+\ldots .$ is

(a) $121(\sqrt{6}+\sqrt{2})$

(b) $243(\sqrt{3}+1)$

(c) $\frac{121}{\sqrt{3}-1}$

(d) $242(\sqrt{3}-1)$

Solution:

(a) $121(\sqrt{6}+\sqrt{2})$

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=\sqrt{2 \times 3^{n-1}}=\sqrt{2}\left(\sqrt{3^{n-1}}\right)$

Now, let $S_{10}$ be the sum of 10 terms of the given series.

Thus, we have:

$S_{10}=\sqrt{2} \sum_{k=1}^{10}\left(\sqrt{3^{(k-1)}}\right)$

$\Rightarrow S_{10}=\sqrt{2}\left(1+\sqrt{3}+\sqrt{3^{2}}+\ldots+\sqrt{3^{9}}\right)$

$\Rightarrow S_{10}=\sqrt{2}\left(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\right)$

$\Rightarrow S_{10 n}=\sqrt{2}\left(\frac{3^{5}-1}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}+1}{\sqrt{3}+1}\right)$

$\Rightarrow S_{10}=\frac{\sqrt{2}}{2}\left(3^{5}-1\right)(\sqrt{3}+1)$

$\Rightarrow S_{10}=\frac{1}{2}(242)(\sqrt{6}+\sqrt{2})$

$\Rightarrow S_{10}=121(\sqrt{6}+\sqrt{2})$