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Question
Find the sum of the following geometric series:
\[\sqrt{2} + \frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}} + . . .\text { to 8 terms };\]
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Solution
Here, a = \[\sqrt{2}\] and r = \[\frac{1}{2}\] .
\[S_8 = a\left( \frac{1 - r^8}{1 - r} \right)\]
\[ = \sqrt{2}\left( \frac{1 - \left( \frac{1}{2} \right)^8}{1 - \frac{1}{2}} \right)\]
\[ = \sqrt{2}\left( \frac{1 - \frac{1}{256}}{\frac{1}{2}} \right)\]
\[ = 2\sqrt{2}\left( \frac{255}{256} \right)\]
\[ = \frac{255\sqrt{2}}{128}\]
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