Formula

Sums of the first $n$ natural numbers, their squares and their cubes

Algebra • Sequence and Series • Summation

Formula

\sum_{r = 1}^{n} r^{3} = [\frac{n ( n + 1 )}{2}]^{2}

\sum_{r = 1}^{n} r = \frac{n ( n + 1 )}{2}

\sum_{r = 1}^{n} r^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

and

\sum_{r = 1}^{n} r^{3} = [\frac{n ( n + 1 )}{2}]^{2}

, so the sum of the first

n

cubes is the square of the sum of the first

n

natural numbers.

How to use it

1Compute $\frac{n ( n + 1 )}{2}$ first: it is wanted in its own right, and its square is the sum of the cubes.
2In the squares formula one of $n$ , $n + 1$ , $2 n + 1$ is always a multiple of $3$ and the pair $n$ , $n + 1$ supplies the factor $2$ , so cancel the $6$ before multiplying out.
3A sum that starts elsewhere is a difference of two of these: $\sum_{r = m}^{n} = \sum_{r = 1}^{n} - \sum_{r = 1}^{m - 1}$ .

Find

1^{2} + 2^{2} + \dots + 7^{2}

Step-by-step

1
$\sum_{r = 1}^{7} r^{2} = \frac{7 \times 8 \times 15}{6}$
Why: Apply the formula with $n = 7$ .
2
$= \frac{840}{6} = 140$

Answer

140

Watch out

All three formulas count from

r = 1

; applying one to a sum that starts elsewhere without subtracting the missing head is the usual slip. That

(\sum r)^{2} = \sum r^{3}

is special to cubes and has no analogue for other powers.

Also known as

sum of the first n natural numbers, sum of squares formula, sum of cubes formula, n(n+1)/2, power sums