Sum of the First n Natural Numbers

The series that gives the sum of the first n natural numbers is defined by $$ s_n = \sum_{k=1}^n k $$

This is one of the simplest and most important series in mathematics. It answers a basic question: what do we get when we add the numbers from 1 up to n?

Let's look at the first few partial sums.

$$ s_1 = 1 \\ s_2 = 1 + 2 = 3 \\ s_3 = 1 + 2 + 3 = 6 \\ s_4 = 1 + 2 + 3 + 4 = 10 \\ s_5 = 1 + 2 + 3 + 4 + 5 = 15 \\ \vdots $$

So the sequence of partial sums begins as follows:

$$ 1 \: , \: 3 \: , \: 6 \: , \: 10 \: , \: 15 \: , \: \dots $$

Gauss's Formula

The problem of summing consecutive integers is traditionally associated with Gauss, who recognized that these sums follow a simple pattern.

Instead of adding the numbers one by one, we can use a direct formula, known as Gauss's formula:

$$ s_n = \sum_{k=1}^n k = \frac{n(n+1)}{2} $$

This formula gives the result immediately, without computing every term. For example:

$$ s_1 = \frac{1(1+1)}{2} = 1 $$

$$ s_2 = \frac{2(2+1)}{2} = 3 $$

$$ s_3 = \frac{3(3+1)}{2} = 6 $$

$$ s_4 = \frac{4(4+1)}{2} = 10 $$

$$ s_5 = \frac{5(5+1)}{2} = 15 $$

Checking a few cases suggests that the formula works, but this is not enough to prove it in general.

$$ s_n = \sum_{k=1}^n k = \frac{n(n+1)}{2} $$

To establish that the formula holds for every natural number, we use mathematical induction.

Base case

We start with n = 1.

$$ P(1) \: : \: \sum_{k=1}^1 k = \frac{1(1+1)}{2} = 1 $$

The statement is true for n = 1.

Inductive hypothesis

Assume that the formula holds for some natural number n.

$$ P(n) \: : \: \sum_{k=1}^n k = \frac{n(n+1)}{2} $$

Inductive step

We now show that it also holds for n + 1.

Add n + 1 to both sides of the hypothesis:

$$ P(n+1) \: : \: \sum_{k=1}^{n+1} k = \frac{n(n+1)}{2} + (n+1) $$

$$ P(n+1) \: : \: \sum_{k=1}^{n+1} k = \frac{n(n+1) + 2(n+1)}{2} $$

$$ P(n+1) \: : \: \sum_{k=1}^{n+1} k = \frac{(n+1)(n+2)}{2} $$

This is exactly the same formula with n replaced by n + 1, so the step is verified.

Note. If we introduce the auxiliary variable z = n + 1, the structure becomes even clearer. $$ P(n+1) \: : \: \sum_{k=1}^{n+1} k = \frac{(n+1)(n+2)}{2} $$ $$ P(n+1) \: : \: \sum_{k=1}^{n+1} k = \frac{(n+1)(n+1+1)}{2} $$ $$ P(z) \: : \: \sum_{k=1}^{z} k = \frac{z(z+1)}{2} \:\:\:\:\: con \: z=n+1 $$

Therefore, the statement holds for n + 1.

By mathematical induction, Gauss's formula is valid for every natural number n.

And so on.