Question

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

• A. $121(\sqrt{6}+\sqrt{2})$
• B. $243(\sqrt{3}+1)$
• C. $\frac{121}{\sqrt{3}-1}$
• D. $242(\sqrt{3}-1)$

Answer 1

The correct option is A

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

$Let{T}_n\text{be the nth term of the given series}\text{Thus, we have:}{T}_n=\sqrt{2\times 3^{n-2}}=\sqrt{2}\left(\sqrt{3^{n-1}}\right)$ Now, let $S_{10}$ be the sum of 10 terms of the given series. Thus, we have: $T_n=\sqrt{2\times 3^{n-1}}=\sqrt{2}\left(\sqrt{3^{n-1}}\right)$ $S_{10}=\sqrt{2}{\sum }_{k=1}^{10}\left(\sqrt{3^{k-1}}\right)\Rightarrow S_{10}=\sqrt{2}(1+\sqrt{3}+\sqrt{3^2}+...+\sqrt{3^9})\Rightarrow S_{10}=\sqrt{2}\left(\frac{\sqrt{3^{10}}-1}{\sqrt{3}-1}\right)\Rightarrow S_{10n}=\sqrt{2}\left(\frac{3^5-1}{\sqrt{3}-1}\right)\left(\frac{\sqrt{3}+1}{\sqrt{3}+1}\right)\Rightarrow S_{10}=\frac{\sqrt{2}}{2}(3^5-1)(\sqrt{3}+1)\Rightarrow S_{10}=\frac{1}{2}\left(242\right)(\sqrt{6}+\sqrt{2})\Rightarrow S_{10}=121(\sqrt{6}+\sqrt{2})$