# Prove $\frac{0}{N} + \frac{1}{N} + \ldots + \frac{q-1}{N} =\frac{q(q-1)}{2N}$

Why this equation holds? Is that by definition from a well known serie?

$$\frac{0}{N} + \frac{1}{N} + \ldots + \frac{q-1}{N} =\frac{q(q-1)}{2N}$$

-
$1+\cdots+n=n(n+1)/2$ –  wj32 Feb 26 '13 at 10:39
Looks like an arithmetic sequence to me. Looks like there's an obvious simplification too: factor out the common term. –  Hurkyl Feb 26 '13 at 12:15

## 2 Answers

First we have $$\frac{1}{N}+\frac{2}{N}+\cdots+\frac{q-1}{N}=\frac{1}{N}\left(1+2+\cdots +(q-1)\right),$$ and then we use that (see this) $$1+2+\cdots + (q-1)=\frac{(q-1)q}{2}.$$

-

Your equation is a result of the following: $1+2+3...+n=\frac12n(n+1)$

We can prove this by considering, $k^2-(k-1)^2=2k-1$ and taking the sum of both sides.

-