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.
 A: 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:
$$\text{Assume } k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k-r}\bigg]$$
$$\text{Then } 10k=\sum_{r=1}^k\bigg[(10-9r)\cdot10^{k+1-r}\bigg]$$
$$\text{So } \sum_{r=1}^{k+1}\bigg[(10-9r)\cdot10^{k+1-r}\bigg]=10k+(10-(9k+9))\cdot10^{(k+1)-(k+1)}$$
$$=10k+(1-9k)\cdot1=k+1 \text{ a.r.}$$
A: An example might help explain the pattern:
$$\begin{align}
12{\color\red3}\times8+{\color\red3}=987
&\implies12{\color\red3}0\times8+{\color\red3}\times10=9870\\
&\implies12{\color\red3}{\color\green4}\times8+{\color\green4}=9870+{\color\green4}\times8+{\color\green4}-{\color\red3}\times10=9870+({\color\green4}-{\color\red3})\times10-{\color\green4}=9870+{\color\yellow6}
\end{align}$$
(My apologies if the colors, in particular the yellow $6$ at the very end, are hard to see.)
A: $$\left\lfloor {10^n\over (1-x)^2} \right\rfloor \cdot 8+n= 9\cdot \left\lfloor {10^n\over (1-x)^2} \right\rfloor -\left\lfloor {10^n\over(1-x)^2} \right\rfloor +n$$
With $x=1$ is what you've observed ( yes I realize the division by 0, just don't know a better way yet to present what the OP sees). The real question though is what makes it work.  
A: This is something I noticed but I'm still thinking about if it means anything:
$$\boxed{1\cdot8+1=9}\\\downarrow$$
$$10\cdot8+10=90$$
$$10\cdot8+18=98$$
$$(10+2)\cdot8+2=98$$
$$\boxed{12\cdot8+2=98}\\\downarrow$$
$$120\cdot8+20=980$$
$$120\cdot8+27=987$$
$$(120+3)\cdot8+3=987$$
$$\boxed{123\cdot8+3=987}\\\downarrow$$
$$1230\cdot8+30=9870$$
$$1230\cdot8+36=9876$$
$$(1230+4)\cdot8+4=9876$$
$$\boxed{1234\cdot8+4=9876}\\\downarrow\\\cdot\\\cdot\\\cdot$$
A: 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.
