Som van machten

$\begin{array}{l} \sum\limits_{k = 0}^n {2^k } = 2^{n + 1} - 1 \\ 2\sum\limits_{k = 0}^n {3^k } = 3^{n + 1} - 1 \\ 4\sum\limits_{k = 0}^n {5^k = 5^{n + 1} - 1} \\ ... \\ \left( {p - 1} \right)\sum\limits_{k = 0}^n {p^k = p^{n + 1} - 1} \\ \end{array}$

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