Question

The sum of 10 terms of the series $\sqrt{2}+\sqrt{6}+\sqrt{18}+$.... is
(a) $121(\sqrt{6}+\sqrt{2})$

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

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

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

Answer 1

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

Let $T_n$ be the nth 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:

$$\begin{gathered} S_{10}=\sqrt{2}\sum _{k=1}^{10}\left(\sqrt{3^{\left(k-1\right)}}\right) \\ \Rightarrow S_{10}=\sqrt{2}\left(1+\sqrt{3}+\sqrt{3^2}+...+\sqrt{3^9}\right) \\ \Rightarrow S_{10}=\sqrt{2}\left(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\right) \\ \Rightarrow S_{10n}=\sqrt{2}\left(\frac{3^5-1}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}+1}{\sqrt{3}+1}\right) \\ \Rightarrow S_{10}=\frac{\sqrt{2}}{2}\left(3^5-1\right)\left(\sqrt{3}+1\right) \\ \Rightarrow S_{10}=\frac{1}{2}\left(242\right)\left(\sqrt{6}+\sqrt{2}\right) \\ \Rightarrow S_{10}=121\left(\sqrt{6}+\sqrt{2}\right) \end{gathered}$$