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I have recently noticed the odd fact that $$123456789\times8+9=987654321$$ So, I started looking into this and I couldn't find anything on it. I decided to figure out a formula for this and what I came up with is $$\sum_{n=1}^{b-1}{(b-2)(b-n)b^{n-1}} = \sum_{n=1}^{b-1}nb^{n-1}$$ for $b=10$. The interesting fact is, that this is true for all positive integer values for b, meaning that $1234...(b-1)\times (b-2)+b-1=(b-1)...4321$ in any base

My question is, why is this true, and is there a proof?

Tbw
  • 995

1 Answers1

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Trivially, the series on the left is summed up to

$$\frac{(b-2) \left(b^b-b^2+b-1\right)}{(b-1)^2}$$

And the one on the right hand side is

$$\frac{(b-2) b^b+1}{(b-1)^2}$$

Their difference is $0$ hence the identity holds.

Enrico M.
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