Summing the First n Natural Numbers

Gautam Sethi

Here is a simple derivation of the sum of the first n natural numbers. In other words,

we want to get an expression for Sn, where

Sn = 1 + 2 + 3 + . . . + (n− 2) + (n− 1) + n

We start with visualizing an n× n square. If n = 10, we have the following square.

As one can see, the required sum Sn is simply a count of the red unit squares. Note

that the remaining unit squares (colored blue) is simply the number Sn−1. Thus, we

have

Sn + Sn−1 ≡ n2 (1)

1

This can be proven more formally as follows.

n2 ≡ n× n

≡ n + n + n + . . . (n times)

≡ [1 + (n− 1)] + [2 + (n− 2)] + [3 + (n− 3)] + · · ·+ [(n− 2) + 2] + [(n− 1) + 1] + n

≡ [1 + 2 + · · ·+ (n− 2) + (n− 1) + n] + [(n− 1) + (n− 2) + . . . + 2 + 1]

≡ [1 + 2 + · · ·+ (n− 2) + (n− 1) + n] + [1 + 2 + · · ·+ (n− 2) + (n− 1)]

≡ Sn + Sn−1

By definition of Sn,

Sn − Sn−1 ≡ n (2)

Therefore we have two equations [equation (1) and equation (2)] in two unknowns,

Sn and Sn−1. Adding the corresponding sides of these two equations, we have

2Sn ≡ n2 + n

≡ n(n + 1)

which implies

Sn ≡n(n + 1)

2

2