123456789 x 8 + 9 = 987654321
Theorem
| \(\displaystyle 1 \times 8 + 1\) | \(=\) | \(\displaystyle 9\) | |||||||||||
| \(\displaystyle 12 \times 8 + 2\) | \(=\) | \(\displaystyle 98\) | |||||||||||
| \(\displaystyle 123 \times 8 + 3\) | \(=\) | \(\displaystyle 987\) | |||||||||||
| \(\displaystyle 1234 \times 8 + 4\) | \(=\) | \(\displaystyle 9876\) | |||||||||||
| \(\displaystyle 12345 \times 8 + 5\) | \(=\) | \(\displaystyle 98765\) | |||||||||||
| \(\displaystyle 123456 \times 8 + 6\) | \(=\) | \(\displaystyle 987654\) | |||||||||||
| \(\displaystyle 1234567 \times 8 + 7\) | \(=\) | \(\displaystyle 9876543\) | |||||||||||
| \(\displaystyle 12345678 \times 8 + 8\) | \(=\) | \(\displaystyle 98765432\) | |||||||||||
| \(\displaystyle 123456789 \times 8 + 9\) | \(=\) | \(\displaystyle 987654321\) |
The above pattern is an instance of the identity:
- $\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$
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 $P \left({n}\right)$ be the proposition:
- $\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$
Basis for the Induction
$P \left({1}\right)$ is the case:
| \(\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^1 j b^{1 - j} + 1\) | \(=\) | \(\displaystyle \left({b - 2}\right) b^0 + 1\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle b - 1\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 1}^1 \left({b - j}\right) b^{1 - j}\) |
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:
- $\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k = \sum_{j \mathop = 1}^k \left({b - j}\right) b^{k - j}$
from which it is to be shown that:
- $\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 1}\right) = \sum_{j \mathop = 1}^{k + 1} \left({b - j}\right) b^{k + 1 - j}$
Induction Step
This is the induction step:
| \(\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + \left({k + 1}\right)\) | \(=\) | \(\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k + 1 - j} + k + 1 + \left({b - 2}\right) \left({k + 1}\right)\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({b - 2}\right) b \sum_{j \mathop = 1}^k j b^{k - j} + \left({b k - b k}\right) + k + 1 + b k - 2 k + b - 2\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle b \left({\left({b - 2}\right) \sum_{j \mathop = 1}^k j b^{k - j} + k}\right) + b - k - 1\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle b \left({\sum_{j \mathop = 1}^k \left({b - j}\right) b^{k - j} }\right) + b - k - 1\) | Induction Hypothesis | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 1}^k \left({b - j}\right) b^{k + 1 - j} + b - \left({k + 1}\right)\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{j \mathop = 1}^{k + 1} \left({b - j}\right) b^{k + 1 - j}\) |
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\displaystyle \left({b - 2}\right) \sum_{j \mathop = 1}^n j b^{n - j} + n = \sum_{j \mathop = 1}^n \left({b - j}\right) b^{n - j}$
$\blacksquare$
