Sum of first \(n\) natural numbers

An Introduction to Real Analysis

2025-09-27

Prove by induction that for all \(n \in \symbb{N}\) \[ \sum_{k = 1}^n k = \frac{n(n + 1)}{2}. \]

Below are two proofs: first by induction, and second by a more intuitive geometric argument.

Proof (First). Since \(\sum_{k = 1}^1 k = 1 = 1(1 + 1) / 2\), the statement \(P(1)\) is true.

Next, assuming that \(P(n)\) is true, \[\begin{align*} \sum_{k = 1}^{n + 1} k &= (n + 1) + \sum_{k = 1}^n k\\[0.75em] &= (n + 1) + \frac{n(n + 1)}{2}\\[0.75em] &= \frac{2(n + 1) + n(n + 1)}{2}\\[0.75em] &= \frac{(n + 1)(2 + n)}{2}\\[0.75em] &= \frac{(n + 1)\left( (n + 1) + 1 \right)}{2}, \end{align*}\] so \(P(n + 1)\) is true. \(\blacksquare\)

Figure 1: Starting with the black dots, add an equal number of white dots. By arranging the dots such that each column has height \(n + 1\), the number of black and white dots is seen to be \(\text{width} \cdot \text{height} = n(n + 1)\). Therefore the number of black dots is \(n(n + 1)/2\).

Proof (Second). Figure 1 gives the geometric inspiration for the second proof, which does not explicitly rely on induction. The figure suggests doubling the sum \(\sum_{k = 1}^n k\) and arranging the two summations in opposite orders, one increasing and the other decreasing, to effectively sum the columns in Figure 1: \[\begin{align*} 2\sum_{k = 1}^n k &= \sum_{k = 1}^n k + \sum_{k = 1}^n k\\[0.75em] &= \sum_{k = 1}^n k + \sum_{k = 1}^n \left( (n + 1) - k \right)\\[0.75em] &= \sum_{k = 1}^n \left( k + (n + 1) - k \right)\\[0.75em] &= \sum_{k = 1}^n (n + 1)\\[0.75em] &= n(n + 1), \end{align*}\] therefore \(\sum_{k = 1}^n k = n(n + 1) / 2\). \(\blacksquare\)