What is the proof of of (N–1) + (N–2) + (N–3) + … + 1= N*(N–1)/2 [closed]

Answer 1

Here's a proof by induction, considering N terms, but it's the same for N - 1:

For N = 0 the formula is obviously true.

Suppose 1 + 2 + 3 + ... + N = N(N + 1) / 2 is true for some natural N.

We'll prove 1 + 2 + 3 + ... + N + (N + 1) = (N + 1)(N + 2) / 2 is also true by using our previous assumption:

1 + 2 + 3 + ... + N + (N + 1) = (N(N + 1) / 2) + (N + 1) = (N + 1)((N / 2) + 1) = (N + 1)(N + 2) / 2.

So the formula holds for all N.