### Sum of integers

is the number of terms times the average of the first term and the last, .

The picture shows that twice the sum, , so .

**General arithmetic progression**

The sum of terms of an arithmetic progression is the average of the first and last terms, times the number of terms.

This diagram shows that is the size of a rectangle of height and width .

### Sum of squares

### Sum of cubes

For the sum of higher powers, see Jacob Bernoulli’s power sum problem

**Manipulating sums**

where is a permutation of the elements of – i.e. the elements can be taken in any order.

Gauss’ boyhood trick of finding can be seen as an application of these 3 laws.

To find the general sum of an arithmetic progression

We can replace by (commutative law):

Add these 2 together. (associative law)

Take the constant outside the sum. (distributive law)

Divide this by to get:

– the average of the first and last terms, , times the number of terms.

**Example of finding a sum**

I wanted to make a video that starts in very slow-motion, and gradually accelerates up to 30 (different) frames a second. Maybe start with 2 images each 1/2 second long (15 frames each), then 3 images 14 frames long, 4×13, 5×12 etc until 14×3, 15 2 frames long, then return to regular 1 image per 1/30 second. How long would this take? It would take:

frames/30ths of a second.

If we represent the numbers rising (the number of images shown for a given number of frames, e.g. initially, 2 images last for 15 frames each) by , then the other is , and the sum is:

These two are the above 2 series for and , with only slight modification:

So the sum can be written as:

with , i.e.

Another way using **Newton’s formula**:

where

and the falling power

The sought value is a term in the series:

We need the term ending in , evidently the 14th term.

Add a first term for , and the values in this case are:

Newton’s formula gives:

Well, what’s the general formula, for starting with a number of images each shown for 30ths of a second? i.e. the sum is then:

If one number from each of these pairs is , the other will be equal to . (Like both numbers always added to 17 in the last example.)

So this time we have

And just to check this formula, plugging in and :

frames.

Knuth et al – Concrete Mathematics