Induction of a Hard Sum Whilst showing proof

Hello
I need help solving this sum, as I know how to solve sums, just not this one since it uses a few things that I havent seen before unfortunately.
The sum is:

n - 1
SUM (n - i) = (n ( n - 1 ) ) / 2
i = 1

Thanks for all the help! :D

Hello,

It's easy to prove:

 n–1
  ∑ (n – i)
 i=1
     = (n–1) + (n–2) + (n–3) + ... + [n–(n–2)] + [n–(n–1)]
     = (n–1) + (n–2) + (n–3) + ... + 2 + 1
     = 1 + 2 + 3 + ... + (n–2) + (n–1) →→→ Which is a known sum
     = n(n – 1)/2

Just in case:
S = 1 + 2 + 3 + ... + (n–3) + (n–2) + (n–1)
S = (n–1) + (n–2) + (n–3) + ... + 3 + 2 + 1
by summing the lines above
2S = (n–1)+1 + (n–2)+2 + (n–3)+3 + ... + 3+(n–3) + 2+(n–2) + 1+(n–1)
2S = n + n + n + ... + n + n + n    (n–1 times)
2S = n(n – 1)
S = n(n – 1)/2

Regards,
Dragon.Jade :-)