Sum of an Arithmetic Series

This was supposedly demonstrated by Gauss, aged eight or something, when he was set a problem by a teacher desperate to get on with something more interesting to add together all the numbers from 1-100. He's supposed to have realised that if you paired up the numbers, they all had the same sum ie 101. It was then a matter of spotting the number of pairs and multiplying the two (5050).

The sum of an arithmetic series is done in a similar way. Think of a series as being:

a₁ + (a₁ + d) + (a₁ + 2d) +......

since d is a constant...

You want to find the sum of the first n terms.

S_n = a₁ + (a₁ + d) + (a₁ + 2d) +....

....(a₁ + (n-1)d)

This is probably quite a few terms (otherwise you wouldn't bother trying to find the sum, would you?).

So the wisest option would be to pair up the terms, like Gauss did, and find the sum of each pair.

The formula is derived from writing the above series out from beginning to n, and the opposite way, and adding the terms.

However, just as Gauss's addition always gave him 101, in our case we will always get 2a + (n-1)d

Since we want the sum to n of the series, there will be n of these pairs. So the sum is n(2a + (n-1)d)

BUT -

We have just added TWO series! Because we wrote it out twice to make adding the terms easier.

Therefore, the final formula will be: n(2a + (n-1)d)/2.


Quite fiddly to prove, but easy enough to use.