123456789 x 8 + 9 = 987654321

Theorem

 $\ds 1 \times 8 + 1$ $=$ $\ds 9$ $\ds 12 \times 8 + 2$ $=$ $\ds 98$ $\ds 123 \times 8 + 3$ $=$ $\ds 987$ $\ds 1234 \times 8 + 4$ $=$ $\ds 9876$ $\ds 12345 \times 8 + 5$ $=$ $\ds 98765$ $\ds 123456 \times 8 + 6$ $=$ $\ds 987654$ $\ds 1234567 \times 8 + 7$ $=$ $\ds 9876543$ $\ds 12345678 \times 8 + 8$ $=$ $\ds 98765432$ $\ds 123456789 \times 8 + 9$ $=$ $\ds 987654321$

The above pattern is an instance of the identity:

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

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:

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

Basis for the Induction

$\map P 1$ is the case:

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

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:

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

from which it is to be shown that:

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

Induction Step

This is the induction step:

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

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

Therefore:

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

$\blacksquare$