# Sum of Cubes (*k*^{3})

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

## Formula

The sum of *k*^{3} 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

1^{3} + 2^{3} + 3^{3} + ... + *k*^{3}

= [*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:

1^{3} + 2^{3} + 3^{3} + ... + *k*^{3} + (*k* + 1)^{3}

= [*k*(*k* + 1)/2]^{2} + (*k* + 1)^{3}.

Make the common factor (*k* + 1)^{2}/4

and add these two fractions.

Adding and subtracting rational expressions

*k*^{2} + 4*k* + 4 = (*k* + 2)^{2}

Factoring a perfect square trinomial

4 = 2^{2}

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*(*k*^{2} + *n*) into (*k*^{3} + *n*⋅*k*).

And rewrite [sum of (*k*^{3} + *n*⋅*k*)]

into [sum of *k*^{3}] + *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 *k*^{3}] = [*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 *n*^{2}(*n* + 1)/4

and add these two fractions.

Adding and subtracting rational expressions