# The Sum of the First n Positive Integers

According to an anecdote, the math genius Carl Friedrich Gauss, added all the numbers from 1 through 100 when he was at a very young age. The sum of the first positive 100 integers is 5050 and Gauss was able to give this sum in less than a minute. True or not, how can we get the sum of all these numbers with admirable speed? Well, it’s not really that hard.

If you want to be a great mathematician someday like Gauss, you must learn to explore problems and look for patterns. Notice that if you add the first $100$ integers, the sum of the largest and the smallest integer is $101$. Also, the sum of the second largest and the second smallest is $101$. This also goes with the third largest and the third smallest and so on. Since there are the first $100$ numbers can be divided into $\frac{100}{2}=50$ pairs, each of which has a sum of $101$, the sum of all the numbers is $50(101) = 5050$. We will use this strategy to generalize. We explore the sum of the first positive n integers.

In adding the first $n$ integers, we have two cases. The first case, if the largest integer is even, like $100$, and the second case is if the largest is odd.

Case 1: Largest integer is even.

If the largest integer is even, we can pair the numbers as shown below.  We can pair $1$ and $n$, $2$ and $n - 1$, $3$ and $n - 2$, and so on.

Notice that each sum is equal to $n + 1$ and there are $\frac{n}{2}$ pairs. Therefore , the sum of all the positive integers from $1$ to $n$ is

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

Case 2:  Largest integer is odd

If the largest integer is odd, we can only pair $n - 1$ integers.  We can pair $1$ and $n - 1$, $2$ and $n - 2$ and $3$ and $n - 3$ and so on.

The sum of each pair above is $n$ and there are $\frac{n - 1}{2}$ pairs. So the sum of the first $n - 1$ positive integers is $\frac{n(n-1)}{2}$. But do not forget that we still have to add the largest digit which is $n$.  So, the sum of all the numbers from $1$ to $n$ where $n$ is odd is

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

As we can see, in any case, the sum of the first $n$ positive integers is $n(n+1)/2$.

The generalization above is not considered as a proof of the sum of all integers. A formal proof called mathematical induction is needed to show that it is true to all positive integers.

## 4 thoughts on “The Sum of the First n Positive Integers”

1. It’s funny, I have used this method several times before without paying much attention to the math behind it. Very interesting that Gauss reportedly did this while very young!