Suppose you begin with a pile of n stones and split this pile into n piles of one stone each by successively splitting a pile of stones into two smaller piles. Each time you split a pile you multiply the number of stones in each of the two smaller piles you form, so that if these piles have $r$ and $s$ stones in them, respectively, you compute $r*s$. Show that no matter how you split the piles, the sum of the products computed at each step equals $n(n − 1)/2$
This is an example in strong induction section. But I think this can be done just use mathematical induction, right?
Basic Step: It's true for $n=2$.
Inductive Step: Assume that it's true for n=k, show that it's also true for n=k+1:
If we have k+1 stones, it can be split into two small piles, $r=k$, $s=1$, we have $r*s=k$, this is one part of our sum, and the rest k stones equals $k(k − 1)/2$, then add k to it, $ k + \frac{k(k − 1)}{2}= \frac{2k}{2} + \frac{k(k − 1)}{2} = \frac{2k+k^2-k}{2} = \frac{k^2+k}{2} = \frac{k(k+1)}{2}$
It's also true for n=k+1.
We proved that it's correct using mathematical induction.
Did I do anything wrong?
