123456789 x 8 + 9 = 987654321

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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:

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

$\blacksquare$