The sum of the first n natural numbers

Submitted by Isoscel on Wed, 09/05/2007 - 20:07.

What is the sum of the first n natural numbers?

Let's look at this problem for n=1, 2, 3, 4, and 5 and calculate the sum:

1=1
1+2=3
1+2+3=6
1+2+3+4=10
1+2+3+4+5=15

What is the formula to calculate this sum:

$S_n=1+2+...+n=?$

The answer is

$S_n=\frac{n(n+1)}{2}$

Here is a calculator that calculates this function for you:

n:

1+2+...+n:

We shall give three different proofs for this formula.

Proof 1:

the numbers put into an rectangle

This is an example for n = 5. We see that we have a big rectangle with the its sides 5 and 5+1. The rectangle has 2(1 + 2 + 3 + 4 + 5) squares inside. So 2(1 + 2+ 3 + 4 + 5) = 5(5+1) and 1 + 2 + 3 + 4 + 5 = $\frac{5(5+1)}{2}$

The same way, for any n, we construct a rectangle with its sides n and n+1 that has 2(1+2+...+n) squares inside.

Proof 2:

$1+2+...+(n-1)+n=S_n$

$(1+2+....+(n-1)+n) + (1+2+....+(n-1)+n) = 2S_n$

(1+2+....+(n-1)+n) +
(n+(n-1)+....+2+1) =
$2S_n$

if we look at the sum above we notice that we have n columns of numbers and that on each column we have two numbers with the sum n+1, so:

$n(n+1) = 2S_n$

$S_n=\frac{n(n+1)}{2}$

Proof 3:

Let's write $S_n=1+2+...+n$ like this: ,

$S_n=\sum_{i=1}^{n}i$

. It's shorter.

We will use induction to prove that:

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

1) For n = 1 it is true that

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

2)We know that

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

is true and we want to prove it for n+1.
We have to prove that:

$S_{n+1}=\sum_{i=1}^{n+1}i=\frac{(n+1)(n+2)}{2}$

if we add n+1 to each side of the next identity:

$S_n=\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$

we have:

$S_{n+1}=S_n+(n+1)=\sum_{i=1}^{n}i+n+1=\frac{n(n+1)}{2}+n+1$

equivalent with:

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

and this leads to:

$S_{n+1}=\sum_{i=1}^{n+1}i=\frac{(n+1)(n+2)}{2}$

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The sum of the first n natural numbers