# Sum of Squares

How to solve the given summation by using the sum of squares (k^{2}) formula: formula, 1 example, and its solution.

## Formula∑_{k = 1}^{n} k^{2}

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

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

This means

1^{2} + 2^{2} + 3^{2} + ... + n^{2} = n(n + 1)(2n + 1)/6.

## Example∑_{k = 1}^{n} k(3k + 1)

k(3k + 1) = 3k^{2} + k

The coefficient 3 gets out from the sigma.

k^{2} is in the sigma.

k is in the sigma.

Summation: How to Solve

Write the coefficient 3.

The sum of k^{2} is

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

The sum of k is

n(n + 1)/2.

Sum of k

So 3⋅n(n + 1)(2n + 1)/6 + n(n + 1)/2.

Cancel the denominator 3

and reduce the numerator 6 to, 6/3, 2.

The common factor of

n(n + 1)(2n + 1)/2 and +n(n + 1)/2

is n(n + 1)/2.

So write

n(n + 1)/2.

Write, n(n + 1)(2n + 1)/2 ÷ n(n + 1)/2,

(2n + 1.

And write, +n(n + 1)/2 ÷ n(n + 1)/2,

+1).

2n + 1 + 1 = 2n + 2

(2n + 2)/2 = (n + 1)

(n + 1)(n + 1) = (n + 1)^{2}

So

n(n + 1)^{2}

is the answer.