# Sum of Cubes (k3)

How to use the sum of cubes formula to solve summation problems: formula, proof, example, and its solution.

## Formula

The sum of k3 as k goes from 1 to n
is [n(n + 1)/2]2.

## Proof

Let's prove this formula
by using mathematical induction.

Show that
'if n = 1, the given statement is true'.

Assume that
'if n = k, the given statement is true'.

So assume that
13 + 23 + 33 + ... + k3
= [k(k + 1)/2]2
is true.

Show that
'if n = k + 1, the given statement is true'.

So add (k + 1)3 on both sides:

13 + 23 + 33 + ... + k3 + (k + 1)3
= [k(k + 1)/2]2 + (k + 1)3.

Make the common factor (k + 1)2/4

k2 + 4k + 4 = (k + 2)2

Factoring a perfect square trinomial

4 = 22

Then [(k + 1)[(k + 1) + 1]/2]2.

So 'if n = k + 1, the given statement is true'.

So, by mathematical induction,
the given statement is true.

## Example

Change k(k2 + n) into (k3 + nk).

And rewrite [sum of (k3 + nk)]
into [sum of k3] + n⋅[sum of k].

(It is k that goes from 1 to n, not n.
So treat n as a constant.)

Basic properties of summation

As as k goes from 1 to n,

[sum of k3] = [n(n + 1)/2]2,
[sum of k] = n(n + 1)/2.

So (given) = [n(n + 1)/2]2 + n⋅[n(n + 1)/2].

Make the common factor n2(n + 1)/4