123456789 x 9 + 10 = 1111111111

Theorem
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:

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:

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

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