# 123456789 x 9 + 10 = 1111111111

## Theorem

 $\ds 1 \times 9 + 2$ $=$ $\ds 11$ $\ds 12 \times 9 + 3$ $=$ $\ds 111$ $\ds 123 \times 9 + 4$ $=$ $\ds 1111$ $\ds 1234 \times 9 + 5$ $=$ $\ds 11111$ $\ds 12345 \times 9 + 6$ $=$ $\ds 111111$ $\ds 123456 \times 9 + 7$ $=$ $\ds 1111111$ $\ds 1234567 \times 9 + 8$ $=$ $\ds 11111111$ $\ds 12345678 \times 9 + 9$ $=$ $\ds 111111111$ $\ds 123456789 \times 9 + 10$ $=$ $\ds 1111111111$

The above pattern is an instance of the identity:

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

where $b = 10$ and $n$ goes from $1$ to $9$.

## Proof

The proof proceeds by induction.

Let $n, b \in \Z_{>0}$, where $b \ge 3$.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:

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

### Basis for the Induction

$\map P 1$ is the case:

 $\ds \paren {b - 1} \sum_{j \mathop = 1}^1 j b^{1 - j} + 2$ $=$ $\ds \paren {b - 1} b^0 + 2$ $\ds$ $=$ $\ds b + 1$ $\ds$ $=$ $\ds \sum_{j \mathop = 0}^1 b^j$

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:

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

from which it is to be shown that:

$\ds \paren {b - 1} \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + k + 2 = \sum_{j \mathop = 0}^{k + 1} b^j$

### Induction Step

This is the induction step:

 $\ds$  $\ds \paren {b - 1} \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + k + 2$ $\ds$ $=$ $\ds \paren {b - 1} \sum_{j \mathop = 1}^k j b^{k + 1 - j} + k + 2 + \paren {b - 1} \paren {k + 1}$ $\ds$ $=$ $\ds \paren {b - 1} b \sum_{j \mathop = 1}^k j b^{k - j} + \paren {b \paren {k + 1} - b \paren {k + 1} } + k + 2 + \paren {b - 1} \paren {k + 1}$ $\ds$ $=$ $\ds b \paren {\paren {b - 1} \sum_{j \mathop = 1}^k j b^{k - j} + k + 1} + 1$ $\ds$ $=$ $\ds b \paren {\sum_{j \mathop = 0}^k b^j} + 1$ Induction Hypothesis $\ds$ $=$ $\ds \sum_{j \mathop = 0}^k b^{j + 1} + 1$ $\ds$ $=$ $\ds \sum_{j \mathop = 1}^{k + 1} b^j + 1$ Translation of Index Variable of Summation: Corollary $\ds$ $=$ $\ds \sum_{j \mathop = 0}^{k + 1} b^j$

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

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

$\blacksquare$