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प्रश्न
Find the sum :
\[\sum^{10}_{n = 1} \left[ \left( \frac{1}{2} \right)^{n - 1} + \left( \frac{1}{5} \right)^{n + 1} \right] .\]
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उत्तर
\[\sum^{10}_{n = 1} \left[ \left( \frac{1}{2} \right)^{n - 1} + \left( \frac{1}{5} \right)^{n + 1} \right]\]
\[ = \sum^{10}_{n = 1} \left( \frac{1}{2} \right)^{n - 1} + \sum^{10}_{n = 1} \left( \frac{1}{5} \right)^{n + 1} \]
\[ = \left\{ 1 + \frac{1}{2} + \frac{1}{4} + . . . + \left( \frac{1}{2} \right)^9 \right\} + \left\{ \frac{1}{5^2} + \frac{1}{5^3} + \frac{1}{5^4} + . . . + \frac{1}{5^{11}} \right\}\]
\[ = 1\left( \frac{1 - \left( \frac{1}{2} \right)^{10}}{1 - \frac{1}{2}} \right) + \frac{1}{25}\left( \frac{1 - \left( \frac{1}{5} \right)^{10}}{1 - \frac{1}{5}} \right) \]
\[ = \left( \frac{2^{10} - 1}{2^9} \right) + \left( \frac{5^{10} - 1}{4 \times 5^{11}} \right)\]
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