# Sum of Squares (*k*^{2})

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

## Formula

The sum of *k*^{2} as *k* goes from 1 to *n*

is *n*(*n* + 1)(2*n* + 1)/6.

## 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^{2} + 2^{2} + 3^{2} + ... + *k*^{2} = *k*(*k* + 1)(2*k* + 1)/6

is true.

Show that

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

So add (*k* + 1)^{2} on both sides:

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

= *k*(*k* + 1)(2*k* + 1)/6 + (*k* + 1)^{2}.

Make the common factor (*k* + 1)/6

and add these two fractions.

Adding and subtracting rational expressions

Factoring a quadratic trinomial

Then (*k* + 1)[(*k* + 1) + 1][2(*k* + 1) + 1]/6.

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

So, by mathematical induction,

the given statement is true.

## Example

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

And rewrite [sum of (3*k*^{2} + *k*)]

into 3⋅[sum of *k*^{2}] + [sum of *k*].

Basic properties of summation

As as *k* goes from 1 to *n*,

[sum of *k*^{2}] = *n*(*n* + 1)(2*n* + 1)/6,

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

So (given) = 3⋅[*n*(*n* + 1)(2*n* + 1)/6] + [*n*(*n* + 1)/2].

Make the common factor *n*(*n* + 1)/2

and add these two fractions.

Adding and subtracting rational expressions

Cancel 2

on both of the numerator and the denominator.

Then (given) = *n*(*n* + 1)^{2}.