The Formula for the Finite Sum of Consecutive Positive Integers

Consider the sum $\sum_{j=1}^{n} j = 1 + 2 + ... + n$ for some $n \in \{1, 2, ... \}$. You are likely already familiar with the finite sum of positive integers formula which is given to be:

(1)

\begin{align} \sum_{j=1}^{n} j = \frac{n(n+1)}{2} \end{align}

The result above can be proven rather easily using mathematical induction. It can also be proven geometrically. We will instead look at a combinatorial proof that instead derives the formula above. We must first look at the following theorem.

Theorem 1: If $n$ and $k$ are nonnegative integers then $\displaystyle{\sum_{j=1}^{n} j^{\underline{k}} = \frac{(n+1)^{\underline{k+1}}}{k+1}}$.

• Proof: Recall from the Sums of Binomial Coefficients page that if $n$ and $k$ are nonnegative integers that satisfy $1 \leq k \leq n$ then $\displaystyle{\sum_{j=1}^{n} \binom{j}{k} = \binom{n+1}{k+1}}$. If we multiply both sides of this equality by $k!$ then we get:

(2)

\begin{align} \quad k! \sum_{j=1}^{n} \binom{j}{k} = k! \binom{n+1}{k+1} \\ \quad \sum_{j=1}^{n} k! \cdot \binom{j}{k} = k! \binom{n+1}{k+1} \\ \quad \sum_{j=1}^{n} k! \cdot \frac{j!}{k! (j - k)!} = k! \cdot \frac{(n+1)!}{(k+1)! \cdot (n+1 - (k+1))!} \\ \quad \sum_{j=1}^{n} \frac{j!}{(j - k)!} = \frac{(n+1)!}{(k + 1)(n - k)!} \\ \quad \sum_{j=1}^{n} j \cdot (j - 1) \cdot ... \cdot (j - k + 1) = \frac{(n+1) \cdot n \cdot ... \cdot (n - k + 1)}{k + 1} \\ \quad \sum_{j=1}^{n} j^{\underline{k}} = \frac{(n+1)^{\underline{k+1}}}{k+1} \quad \blacksquare \end{align}

Corollary 1: If $n$ is a positive integer then $\sum_{j=1}^{n} j = \frac{n(n+1)}{2}$.

• Proof: From Theorem 1 we set $k = 1$ to get:

(3)

\begin{align} \quad \sum_{j=1}^{n} j^{\underline{1}} = \frac{(n+1)^{\underline{2}}}{2} \\ \quad \sum_{j=1}^{n} j = \frac{(n+1)n}{2} \quad \blacksquare \end{align}

From Theorem 1 we can also generate a formula for the sum of squares as we demonstrate in the following corollary.

Corollary 2: If $n$ is a positive integer then $\sum_{j=1}^{n} j = \frac{n(n+1)(2n+1)}{6}$.

• Proof: From Theorem 1 we set $k = 2$ to get:

(4)

\begin{align} \quad \sum_{j=1}^{n} j^{\underline{2}} = \frac{(n+1)^{\underline{3}}}{3} \\ \quad \sum_{j=1}^{n} j(j-1) = \frac{(n+1) \cdot n \cdot (n - 1)}{3} \\ \quad \sum_{j=1}^{n} j^2 - \sum_{j=1}^{n} j = \frac{(n+1) \cdot n \cdot (n - 1)}{3} \end{align}

• We apply Corollary 1 to get:

(5)

\begin{align} \quad \sum_{j=1}^{n} j^2 - \frac{n(n+1)}{2} = \frac{(n+1) \cdot n \cdot (n - 1)}{3} \\ \quad \sum_{j=1}^{n} j^2 = \frac{(n+1) \cdot n \cdot (n - 1)}{3} + \frac{n(n+1)}{2} \\ \quad \sum_{j=1}^{n} j^2 = \frac{2 \cdot (n + 1) \cdot n \cdot (n - 1) + 3 n(n+1)}{6} \\ \quad \sum_{j=1}^{n} j^2 = \frac{n(n+1)[2(n - 1) +3]}{6} \\ \quad \sum_{j=1}^{n} j^2 = \frac{n(n+1)(2n + 1)}{6} \quad \blacksquare \end{align}

Formulas for the sum of consecutive positive cubes, etc… can be derived similarly.