Basic Rhythms

Saul Foresta explained as follows why this pattern holds in the decimal system and in other number systems using bases other than base $10$:

I generalized the problem for any base $n$ and any number of digits $r$ where $r$ can be anywhere from $1$ to $(n - 1)$.

Then after rewriting both sides of the equality given in the problem using sigma notation I arrived at the following:

$$ {\sum_{k=1}^r (n-k)n^{r-k}} = {{(n-2)\sum_{k=1}^r kn^{r-k} + r}} $$

In each summation $k$ stands for the $k$th digit of the number we're dealing with, reading from left to right. For example, in the number $9876$, $k$ ranges from $1$-$4$, where $9$ is $k=1$, $8$ is $k=2$, and so on.

So all I need to do in order to prove that this pattern holds is show that the left side of this equality does indeed equal the right side. Taking the terms like $8 \times123456789$, that is

$$ (n-2)\sum_{k=1}^r kn^{r-k} $$

over to the left hand side, we will prove that this expression is equal to $r$.

[(n-1)n ^r-1 + (n-2)n ^r-2 + (n-3)n ^r-3 + ... + (n-r)] - (n-2)[n ^r-1 + 2n ^r-2 + 3n ^r-3 + ... + r] =
[n ^r - n ^r-1 + n ^r-1 - 2n ^r-2 + n ^r-2 - 3n ^r-3 + ...+ n - r] - [n ^r - 2n ^r-1 + 2n ^r-1 - 2.2n ^r-2 + ... + (n-2)r]

The coefficient of $n^{r-k}$ on this left hand side is $[1-k] - [k+1-2k] = 0$ for

$ ( 1 \le k \le r-1) $

The coefficient of $n^0$ is $[-r] - [-2(r)] = r$ and hence this expression is equal to $r$ as required.