# 123456789 x 9 + 10 = 1111111111

## Theorem

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

The above pattern is an instance of the identity:

$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$

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 $P \left({n}\right)$ be the proposition:

$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$

### Basis for the Induction

$P \left({1}\right)$ is the case:

 $\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^1 j b^{1 - j} + 2$ $=$ $\displaystyle \left({b - 1}\right) b^0 + 2$ $\displaystyle$ $=$ $\displaystyle b + 1$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^1 b^j$

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

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

So this is the induction hypothesis:

$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k + 1 = \sum_{j \mathop = 0}^k b^j$

from which it is to be shown that:

$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 2}\right) = \sum_{j \mathop = 0}^{k + 1} b^j$

### Induction Step

This is the induction step:

 $\displaystyle$  $\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 2}\right)$ $\displaystyle$ $=$ $\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^k j b^{k + 1 - j} + \left({k + 2}\right) + \left({b - 1}\right) \left({k + 1}\right)$ $\displaystyle$ $=$ $\displaystyle \left({b - 1}\right) b \sum_{j \mathop = 1}^k j b^{k - j} + \left({b \left({k + 1}\right) - b \left({k + 1}\right)}\right) + \left({k + 2}\right) + \left({b - 1}\right) \left({k + 1}\right)$ $\displaystyle$ $=$ $\displaystyle b \left({\left({b - 1}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k + 1}\right) + 1$ $\displaystyle$ $=$ $\displaystyle b \left({\sum_{j \mathop = 0}^k b^j}\right) + 1$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^k b^{j + 1} + 1$ $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 1}^{k + 1} b^j + 1$ Translation of Index Variable of Summation: Corollary $\displaystyle$ $=$ $\displaystyle \sum_{j \mathop = 0}^{k + 1} b^j$

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

Therefore:

$\displaystyle \left({b - 1}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$

$\blacksquare$