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:
- $\ds \paren {b - 1} \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$
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 $\map P n$ be the proposition:
- $\ds \paren {b - 1} \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$
Basis for the Induction
$\map P 1$ is the case:
\(\ds \paren {b - 1} \sum_{j \mathop = 1}^1 j b^{1 - j} + 2\) | \(=\) | \(\ds \paren {b - 1} b^0 + 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 0}^1 b^j\) |
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:
- $\ds \paren {b - 1} \sum_{j \mathop = 1}^k j b^{k - j} + k + 1 = \sum_{j \mathop = 0}^k b^j$
from which it is to be shown that:
- $\ds \paren {b - 1} \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + k + 2 = \sum_{j \mathop = 0}^{k + 1} b^j$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds \paren {b - 1} \sum_{j \mathop = 1}^{k + 1} j b^{k + 1 - j} + k + 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {b - 1} \sum_{j \mathop = 1}^k j b^{k + 1 - j} + k + 2 + \paren {b - 1} \paren {k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {b - 1} b \sum_{j \mathop = 1}^k j b^{k - j} + \paren {b \paren {k + 1} - b \paren {k + 1} } + k + 2 + \paren {b - 1} \paren {k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b \paren {\paren {b - 1} \sum_{j \mathop = 1}^k j b^{k - j} + k + 1} + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b \paren {\sum_{j \mathop = 0}^k b^j} + 1\) | 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 $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \paren {b - 1} \sum_{j \mathop = 1}^n j b^{n - j} + n + 1 = \sum_{j \mathop = 0}^n b^j$
$\blacksquare$