Question

Suppose you begin with a pile of n stones and split this pile into n piles of one stone each by successively split- ting 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 rs. Show that no matter how you split the piles, the sum of the products computed at each step equals n(n - 1)/2.

Answer 1

Proof by induction:

If you have one stone, you do no splitting, and 1*0/2 = 0, which is correct.
If you have two stones, you have to split into two piles of size 1.
1 * 1 = 1. n(n-1)/2 = 2(2-1)/2 = 2*1/2 = 2/2 = 1. Correct.
Assume for n = k.
Show for n = k + 1.
When you split the pile of size k+1, you split this into 2 piles of size x and k+1-x, where 1 <= x <= k
Thus, your first product is x(k+1 -x)
By assumption, the pile of size x, when broken up, will yield a product of
x(x-1)/2 and the pile of size k+1-x will yield a pile of size (k+1-x)(k+1-x-1)/2
x(k+1 -x) + x(x-1)/2 + (k+1-x)(k+1-x-1)/2 = (noting that k+1-x-1 = k - x)
xk + x - x2 + x2/ 2 - x/2 + k2 /2 -kx/2 + k/2 - x/2 -kx/2 + x2/ 2 =
k2 /2 + k/2 + xk -kx/2 -kx/2 + x - x/2 - x/2 - x2 + x2/ 2 + x2/ 2 =
k2 /2 + k/2 =
(k+1)(k)/2 =
(k+1)(k+1-1)/2 which is what we were trying to prove.