Question

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 rs. Show that no matter how you split the piles, the sum of the products computed at each step equals (n(n-1))/2 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 rs. Show that no matter how you split the piles, the sum of the products computed at each step equals (n(n-1))/2 @Mathematics

Answer 1

So I tried this question and made a random n such that n=5 and it was true that it does equal (n(n-1))/2 but I don't see how I can prove this. Any help would be very much appreciated!!

Answer 2

there r n stones in the begenning,
and say i take one out,the product is
1(n-1)

now if i take anothe out, i will have
(n-2)2

similarly taking this for n such processe i wil have

(n-1)1 + (n-2)2 + (n-3)3 ...............1(n-1)

Answer 3

after opening the brackets, areid ,u will get

{n+2n+3n+4n......n*n/2....2n +n} -[1+4+9+16.....n*n]