Repunit Integer as Product of Base - 1 by Increasing Digit Integer

Theorem

$\ds 9 \times 1 + 2$

$=$

$\ds 11$

$\ds 9 \times 12 + 3$

$=$

$\ds 111$

$\ds 9 \times 123 + 4$

$=$

$\ds 1111$

$\ds 9 \times 1234 + 5$

$=$

$\ds 11111$

$\ds 9 \times 12345 + 6$

$=$

$\ds 11111$

$\ds $

$\ldots$

$\ds $

That is:

$\ds 9 \sum_{j \mathop = 0}^n \paren {n - j} 10^j + n + 1 = \sum_{j \mathop = 0}^n 10^j$

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

A specific instance of the general result:

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

where $b = 10$.

$\blacksquare$