Here we will prove some special sums that may appear when evaluating integrals using Riemann sums. These do not always appear, but it is important to recognize them for substitution.

# Proof 1 (By Induction): 1 + 2 + 3 + … + n = n(n + 1)/2

We're going to first prove this by the process of mathematical induction. For n ≥ 1, let S(n) be the statement that $1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}$. We first verify this with the base step that n = 1. $1 = \frac{1(1+1)}{2}$, so S(1) is true. Now assume that for some integer k ≥ 1, S(k) is true, that is assume $1 + 2 + 3 + ... + k = \frac{k(k+1)}{2}$ is true. We want to verify S(k+1), or show that $S(k+1): 1 + 2 + 3 + ... + (k+1) = \frac{(k+1)[(k+1) + 1)}{2}$ is true. Hence we want to show that S(k) implies S(k+1), or S(k) → S(k+1).

(1)Therefore S(k) → S(k+1). Hence for all n ≥ 1, S(n) is true.

## Proof 1 (Direct): 1 + 2 + 3 + … + n = n(n + 1)/2

This is perhaps the more preferable proof if you have just came from reading about Calculus and do not fully understand proof by induction. First we will let S(n) = 1 + 2 + 3 + … + n. Also note that S(n) = n + (n - 1) + (n - 2) + … + 1. Let's add both of these expressions together to get 2S(n) = 1 + 2 + 3 + … + n + n + (n-1) + (n - 2) + … + 1. Rearranging the terms we obtain 2S(n) = (1 + n) + (1 + n) + … + (1 + n). In fact, there are a finite sum of (1 + n)'s, n to be exact. Hence 2S(n) = n(1 + n) = n(n + 1). Dividing both sides by 2 we obtain that S(n) = n(n + 1)/2, or rather 1 + 2 + 3 + … + n = n(n + 1)/2