Sum of Geometric Progression/Examples/One Seventh from 1 to n

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Theorem

$\displaystyle \sum_{j \mathop = 0}^n \dfrac 1 {7^j} = \frac 7 6 \left({1 - \frac 1 {7^{n + 1} } }\right)$


Proof

\(\displaystyle \displaystyle \sum_{j \mathop = 0}^n \dfrac 1 {7^j}\) \(=\) \(\displaystyle \frac {1 - \frac 1 7^{n + 1} } {1 - \frac 1 7}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {7 - 7 \frac 1 7^{n + 1} } {7 - 1}\) multiplying top and bottom by $7$

Hence the result.

$\blacksquare$


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