Repunit Integer as Product of Base - 1 by Increasing Digit Integer/General Result

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Theorem

Let $b \in \Z_{>1}$.

Then:

$\ds \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j} b^j + n + 1 = \sum_{j \mathop = 0}^n b^j$


Proof

\(\ds \) \(\) \(\ds \paren {b - 1} \sum_{j \mathop = 0}^n \paren {n - j} b^j + n + 1\)
\(\ds \) \(=\) \(\ds n \paren {b - 1} \sum_{j \mathop = 0}^n b^j - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1\)
\(\ds \) \(=\) \(\ds n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1} \sum_{j \mathop = 0}^n j b^j + n + 1\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds n \paren {b - 1} \frac {b^{n + 1} - 1} {b - 1} - \paren {b - 1} \frac {n b^{n + 2} - \paren {n + 1} b^{n + 1} + b} {\paren {b - 1}^2} + n + 1\) Sum of Sequence of Power by Index
\(\ds \) \(=\) \(\ds \frac {n \paren {b - 1} \paren {b^{n + 1} - 1} - \paren {n b^{n + 2} - \paren {n + 1} b^{n + 1} + b} + \paren {b - 1} n + \paren {b - 1} } {b - 1}\) elementary simplification
\(\ds \) \(=\) \(\ds \frac {n b^{n + 2} - n b^{n - 1} - n b + n - n b^{n + 2} + n b^{n - 1} + b^{n + 1} - b + n b - n + b - 1} {b - 1}\) multiplying out
\(\ds \) \(=\) \(\ds \frac {b^{n + 1} - 1} {b - 1}\) simplification
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 0}^n b^j\) Sum of Geometric Sequence

$\blacksquare$


Sources

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