Sums of Squares and Cubes

The whole numbers

$\sum r$Σr, read "the sum of r from r = 1 to n", means 1 + 2 + 3 + … + n. Its closed formula is $\sum r=\frac{n\left(n+1\right)}{2}$Σr = n(n + 1)/2. For n = 4, 1 + 2 + 3 + 4 = 10, and 4 × 5 ÷ 2 = 10.

Write the sum forwards and backwards and add the two lines column by column: 1 + n, 2 + (n − 1), 3 + (n − 2), and so on. Every column adds to n + 1, and there are n columns, so twice the sum is n(n + 1).

The staircase 1 + 2 + 3 + 4 and an upside-down copy. Slide the copy in: the two make a 4 by 5 rectangle of 20 blocks, so one staircase is 10.

The squares

$\sum r^2=1^2+2^2+\cdots +n^2$Σr² = 1² + 2² + … + n² has its own formula: $\sum r^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$Σr² = n(n + 1)(2n + 1)/6. For n = 3, 1 + 4 + 9 = 14, and 3 × 4 × 7 ÷ 6 = 84 ÷ 6 = 14. For n = 4, 1 + 4 + 9 + 16 = 30, and 4 × 5 × 9 ÷ 6 = 30.

Induction proves it for every n. At n = 1 both sides are 1, since 1 × 2 × 3 ÷ 6 = 1. Suppose it holds for n, and add the next square, ${\left(n+1\right)}^2$(n + 1)²: $\frac{n\left(n+1\right)\left(2n+1\right)}{6}+{\left(n+1\right)}^2=\frac{\left(n+1\right)\left(n\left(2n+1\right)+6\left(n+1\right)\right)}{6}=\frac{\left(n+1\right)\left(2n^2+7n+6\right)}{6}=\frac{\left(n+1\right)\left(n+2\right)\left(2n+3\right)}{6}$n(n + 1)(2n + 1)/6 + (n + 1)² = (n + 1)(n(2n + 1) + 6(n + 1))/6 = (n + 1)(2n² + 7n + 6)/6 = (n + 1)(n + 2)(2n + 3)/6. That is the formula with n + 1 in place of n.

For large n the formula is close to $\frac{n^3}{3}$n³/3, the area under $y=x^2$y = x² from 0 to n. Multiplied out, $\sum r^2=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$Σr² = n³/3 + n²/2 + n/6, so the sum is always a little more than that area.

The totals 1, 5, 14, 30, 55, 91 for n = 1 to 6, with the vertical scale squeezed. The gold curve $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$n(n + 1)(2n + 1)/6 passes through every one; the dashed curve $\frac{n^3}{3}$n³/3 runs just below them.

The cubes

The cubes add to the square of the plain sum: $\sum r^3={\left(\frac{n\left(n+1\right)}{2}\right)}^2$Σr³ = (n(n + 1)/2)². For n = 3, 1 + 8 + 27 = 36, which is $6^2$6², and 1 + 2 + 3 = 6. For n = 4, $1+8+27+64=100=10^2$1 + 8 + 27 + 64 = 100 = 10².

Here is why. Write $T_n=\frac{n\left(n+1\right)}{2}$Tₙ = n(n + 1)/2 for the nth triangle number. Going from $T_{n-1}$Tₙ₋₁ to $T_n$Tₙ adds n, so $T_n-T_{n-1}=n$Tₙ − Tₙ₋₁ = n, and the two add to $n^2$n², since $\frac{\left(n-1\right)n}{2}+\frac{n\left(n+1\right)}{2}=n^2$(n − 1)n/2 + n(n + 1)/2 = n². So $T_n^2-T_{n-1}^2=\left(T_n-T_{n-1}\right)\left(T_n+T_{n-1}\right)=n\times n^2=n^3$Tₙ² − Tₙ₋₁² = (Tₙ − Tₙ₋₁)(Tₙ + Tₙ₋₁) = n × n² = n³: each cube is the step from one squared triangle number to the next. For n = 3, $6^2-3^2=36-9=27$6² − 3² = 36 − 9 = 27.

Adding the steps from the start, $1^2$1², $3^2$3², $6^2$6², $10^2$10², … climb by 1, 8, 27, 64, …, so the sum of the first n cubes is the square of the nth triangle number.

Building other sums

A sum splits term by term, and a constant factor comes out in front: $\sum \left(r^2+2r\right)=\sum r^2+2\sum r=\frac{n\left(n+1\right)\left(2n+1\right)}{6}+n\left(n+1\right)$Σ(r² + 2r) = Σr² + 2Σr = n(n + 1)(2n + 1)/6 + n(n + 1). Taking out $\frac{n\left(n+1\right)}{6}$n(n + 1)/6 gives $\frac{n\left(n+1\right)\left(2n+1+6\right)}{6}=\frac{n\left(n+1\right)\left(2n+7\right)}{6}$n(n + 1)(2n + 1 + 6)/6 = n(n + 1)(2n + 7)/6. For n = 3 the terms are 3, 8 and 15, which add to 26, and 3 × 4 × 13 ÷ 6 = 26.

In the same way $\sum r\left(r+1\right)=\sum r^2+\sum r=\frac{n\left(n+1\right)\left(n+2\right)}{3}$Σr(r + 1) = Σr² + Σr = n(n + 1)(n + 2)/3. For n = 4: 2 + 6 + 12 + 20 = 40, and 4 × 5 × 6 ÷ 3 = 40.

A constant adds n times: $\sum 1=n$Σ1 = n, so $\sum 3=3n$Σ3 = 3n, not 3. That is how $\sum \left(2r-1\right)=2\sum r-n=n\left(n+1\right)-n=n^2$Σ(2r − 1) = 2Σr − n = n(n + 1) − n = n², the sum of the first n odd numbers.

A sum that does not start at 1 is a difference of two sums from 1. Write S(n) for the sum of the first n squares. The squares from $11^2$11² to $20^2$20² add to S(20) − S(10): 20 × 21 × 41 ÷ 6 − 10 × 11 × 21 ÷ 6 = 2870 − 385 = 2485.

The usual mistakes

Using $\sum r$Σr for $\sum r^2$Σr². For n = 4 that gives 10 instead of 30.

Squaring the sum for $\sum r^2$Σr², or cubing it for $\sum r^3$Σr³. For n = 3, ${\left(\sum r\right)}^2$(Σr)² is 36, not 14, and ${\left(\sum r\right)}^3$(Σr)³ is 216, not 36. The cubes add to the square of $\sum r$Σr; the squares do not.

Splitting a sum of products into a product of sums. $\sum \left(r\times r\right)$Σ(r × r) is not $\sum r\times \sum r$Σr × Σr: for n = 3 that is 14 against 36.

Writing $\sum 3$Σ3 as 3. Each of the n terms is 3, so the sum is 3n.

Subtracting the wrong sum for a range. The squares from 11 to 20 are S(20) − S(10); S(20) − S(11) leaves out $11^2$11².

A stack of square layers

In the application below, layer k of a stack holds $k^2$k² pipes, so a stack of n layers holds $\sum r^2$Σr² pipes, and the formula replaces adding the layers one by one.