Proofs Of Special Sums By Induction

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)
\begin{align} \mathbf{LHS \: of \: S(k + 1):} \: 1 + 2 + 3 + ... + (k + 1) \\ = 1 + 2 + 3 + ... + k + (k +1) \\ = \frac{k(k+1)}{2} + (k + 1) \\ = \frac{k(k+1)}{2} + \frac{2(k + 1)}{2} \\ = \frac{k(k+1) + 2(k + 1)}{2} \\ = \frac{(k+1)(k + 2)}{2} \\ = \frac{(k+1)[(k+1) + 1]}{2} \\ \end{align}

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

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