Question

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

 

Options

• \[121 (\sqrt{6} + \sqrt{2})\]

• \[243 (\sqrt{3} + 1)\]

• \[\frac{121}{\sqrt{3} - 1}\]

• \[242 (\sqrt{3} - 1)\]

Answer 1

\[121 (\sqrt{6} + \sqrt{2})\]

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:

\[S_{10} = \sqrt{2} \sum^{10}_{k = 1} \left( \sqrt{3^\left( k - 1 \right)} \right)\]

\[ \Rightarrow S_{10} = \sqrt{2}\left( 1 + \sqrt{3} + \sqrt{3^2} + . . . + \sqrt{3^9} \right)\]

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

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

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

\[ \Rightarrow S_{10} = \frac{1}{2}\left( 242 \right)\left( \sqrt{6} + \sqrt{2} \right)\]

\[ \Rightarrow S_{10} = 121 \left( \sqrt{6} + \sqrt{2} \right)\]