6

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.

Alex
  • 61
  • 3
    Note that $\Sigma_1 \div \Sigma_2 = \frac1{B-2}$ is equivalent to $(B-2)\Sigma_1 = \Sigma_2$, so you might find it easier to work on trying to prove the latter. – MPW Apr 13 '20 at 15:04
  • It will just involve sums of geometric series, except for the term $\sum_i iB^i$. Do you know how to evaluate this? – saulspatz Apr 13 '20 at 15:14

1 Answers1

3

(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.

Brian M. Scott
  • 616,228