123456789 x 9 + 10 = 1111111111

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Theorem

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


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:

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

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:

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

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


Therefore:

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

$\blacksquare$