How to make sense of a weird Summation? I have been toying with some sums, as I saw that $123456789\div987654312 = \frac{1}{8}$. Which led me to wonder if this can be generalized to other bases. I.E. in base 4 it would be $(1*4^{2}+2*4^{1}+3*4^{0})\div(3*4^{2}+1*4^{1}+2*4^{0})=\frac{1}{2}$. So this led me to the generalization of, if B is base $$\left(\sum^{B-1}_{i=0}(B-1-i)*B^{i}\right)\div\left(\left(\sum^{B-2}_{i=2}(i+1)*B^{i}\right)+B + 2\right)=\frac{1}{B-2}$$
I've tested many values on my calculator and I've confirmed for B = 3 to B = 52. If anyone smarted than me could either explain this to me or at least point me a direction to understand. 
 A: (This has hung around unanswered for more than long enough.)
I’ll rewrite the equation like this
$$\frac{\sum\limits_{i=0}^{b-1}(b-1-i)b^i}{b+2+\sum\limits_{i=2}^{b-2}(i+1)b^i}=\frac1{b-2}$$
and (as MPW suggested in the comments) get rid of the fractions to produce the following equation, which is equivalent for $b\ne 2$:
$$(b-2)\sum_{i=0}^{b-1}(b-1-i)b^i=b+2+\sum_{i=2}^{b-2}(i+1)b^i\,.\tag{1}$$
We can rewrite the lefthand side as
$$\begin{align*}
(b-2)\sum_{i=0}^{b-1}(b-1-i)b^i&=(b-2)\sum_{i=1}^bb^i-(b-2)\sum_{i=0}^{b-1}(i+1)b^i\\
&=(b-2)\frac{b^{b+1}-b}{b-1}-(b-2)\sum_{i=0}^{b-1}(i+1)b^i\\
&=\frac{b(b-2)(b^b-1)}{b-1}-(b-2)\sum_{i=0}^{b-1}(i+1)b^i
\end{align*}$$
and the righthand side as
$$\begin{align*}
b+2+\sum_{i=2}^{b-2}(i+1)b^i&=b+2+\sum_{i=0}^{b-1}(i+1)b^i-(1+2b+b^b)\\
&=1-b-b^b+\sum_{i=0}^{b-1}(i+1)b^i\,,
\end{align*}$$
to make the summations match. That will allow us to combine them on one side of the equation, transforming $(1)$ into the equivalent equation
$$\begin{align*}
\frac{b(b-2)(b^b-1)}{b-1}&=1-b-b^b+\sum_{i=0}^{b-1}(i+1)b^i+(b-2)\sum_{i=0}^{b-1}(i+1)b^i\\
&=1-b-b^b+(b-1)\sum_{i=0}^{b-1}(i+1)b^i\,.\tag{2}
\end{align*}$$
There are a number of ways to evaluate that summation; here is a very elementary one that requires only knowing how to reverse the order of a double summation:
$$\begin{align*}
\sum_{i=0}^{b-1}(i+1)b^i&=\sum_{i=0}^{b-1}\left(\sum_{j=0}^i1\right)b^i\\
&=\sum_{i=0}^{b-1}\sum_{j=0}^ib^i\\
&=\sum_{j=0}^{b-1}\sum_{i=j}^{b-1}b^i\\
&=\sum_{j=0}^{b-1}\frac{b^b-b^j}{b-1}\\
&=\frac1{b-1}\left(b^{b+1}-\sum_{j=0}^{b-1}b^j\right)\\
&=\frac1{b-1}\left(b^{b+1}-\frac{b^b-1}{b-1}\right)\,.
\end{align*}$$
We can now simplify $(2)$ to
$$\frac{b(b-2)(b^b-1)}{b-1}=1-b-b^b+b^{b+1}-\frac{b^b-1}{b-1}$$
and thence, after multiplication by $b-1$, to
$$b(b-2)(b^b-1)=2b-b^2-2b^{b+1}+b^{b+2}\,;$$
this is still equivalent to $(1)$, since $b\ne 0$, and is easily verified.
