Why does this pattern occur: $123456789 \times 8 + 9 = 987654321$

I came across the following:

$\begin{align} 1 \times 8 + 1 &= 9 \\ 12 \times 8 + 2 & = 98 \\ 123 \times 8 + 3 & = 987 \\ 1234 \times 8 + 4 & = 9876 \\ 12345 \times 8 + 5 & = 98765 \\ 123456 \times 8 + 6 & = 987654 \\ 1234567 \times 8 + 7 & = 9876543 \\ 12345678 \times 8 + 8 & = 98765432 \\ 123456789 \times 8 + 9 & = 987654321. \\ \end{align}$

I'm looking for an explanation for this pattern. I suspect that there is some connection to the series $\frac{1}{(1 - x)^2} = 1 + 2x + 3x^2 + \cdots$.

This post asks the same question but has no answers posted.

Solution 1:

If I consider the equations you provide with your "ideas so far":

\begin{align} 1 \times 9 + 1 &= 10 \\ 12 \times 9 + 2 & = 110 \\ 123 \times 9 + 3 & = 1110 \\ \vdots\\ 123456789 \times 9 + 9 & = 1111111110, \\ \end{align}

The first equation being true, this system is equivalent to the system composed of their successive differences all of them having the common pattern :

$$\underbrace{11...1}_{k \ \text{digits}} \times 9 + 1 = 10^k$$

which is an (almost) evident fact.

Solution 2:

Rewritten in sum form, your equations become:

$$\bigg(\sum_{r=1}^n8r\cdot10^{n-r}\bigg)+n=\bigg(\sum_{r=1}^n(10-r)\cdot10^{n-r}\bigg)$$ for $n\in\Bbb N\cap[1,9]$

Subtracting the RHS gives:

$$n=\sum_{r=1}^n\bigg[(10-9r)\cdot10^{n-r}\bigg]$$

We prove this via induction:

$$\text{Assume } k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k-r}\bigg]$$ $$\text{Then } 10k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k+1-r}\bigg]$$ $$\text{So } \sum_{r=1}^{k+1}\bigg[(10-9r)\cdot10^{k+1-r}\bigg]=10k+(10-(9k+9))\cdot10^{(k+1)-(k+1)}$$ $$=10k+(1-9k)\cdot1=k+1 \text{ a.r.}$$

Solution 3:

Ideas so far:

Adding another $123 \cdots$ to both sides yields the following equivalent series equations: \begin{align} 1 \times 9 + 1 &= 10 \\ 12 \times 9 + 2 & = 110 \\ 123 \times 9 + 3 & = 1110 \\ \vdots\\ 123456789 \times 9 + 9 & = 1111111110, \\ \end{align} so it suffices to prove that the above pattern holds. We note that for $n = 1,\dots,9$, we can write the first number on the LHS of each equation as $$ 10^{n-1} \cdot (1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)}). $$ Let $M = 1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)}$. We have $$ \begin{align} M &= 1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)} \\ & = (1 + 2 \cdot 10^{-1} + \cdots + n \cdot 10^{-(n-1)} + \cdots) - ((n+1) \cdot 10^{-n} + (n+2) \cdot 10^{-(n+1)} + \cdots) \\ & = \frac{1}{(1 - 10^{-1})^2} - ((n+1) \cdot 10^{-n} + (n+2) \cdot 10^{-(n+1)} + \cdots). \end{align} $$ Let $N = (n+1) \cdot 10^{-n} + (n+2) \cdot 10^{-(n+1)} + \cdots$. We can rewrite this as $$ \begin{align} M &= \sum_{k=n+1}^\infty k\cdot 10^{-(k-1)} = \sum_{k=1}^\infty (k+n)\cdot 10^{-(k+n-1)} \\ & = \sum_{k=1}^\infty k \cdot 10^{-(k+n-1)} + n \cdot \sum_{k=1}^\infty \cdot 10^{-(k+n-1)} \\ & = 10^{-n} \cdot \sum_{k=1}^\infty k \cdot 10^{-(k-1)} + n \cdot 10^{-n} \cdot \sum_{k=1}^\infty \cdot 10^{-(k-1)} \\ & = 10^{-n} \frac{1}{(1 - 10^{-1})^2} + n \cdot 10^{-n} \cdot \frac{1}{1 - 10^{-1}} \\ & = 10^{-n} \cdot \frac{1 + n \cdot(1 - 10^{-n})}{(1 - 10^{-1})^2} \end{align} $$ That is, we have $$ M = \frac{1}{(1 - 10^{-1})^2} - N = \frac{1 - 10^{-n}(1 + n \cdot(1 - 10^{-n}))}{(1 - 10^{-1})^2}. $$ With that, we can rewrite the LHS of the equation as $$ \begin{align} 10^{n-1}M + n &= \frac{10^{n-1} - 10^{-1}(1 + n \cdot(1 - 10^{-n}))}{(1 - 10^{-1})^2} + n \\ & = \frac{10^{n-1} - 10^{-1}(1 + n \cdot(1 - 10^{-n})) + n\cdot (1 - 10^{-1})^2}{(1 - 10^{-1})^2} \end{align} $$