Core Pure · Series · Year 1

Sums of Squares & Cubes

A telescoping trick extends the natural-number formula to squares and cubes — and the cube-sum formula hides a beautifully unexpected pattern.

17 min read 5 worked examples

The sum of squares

Theorem 9.1

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

ProofTelescoping method using (r+1)³ − r³

Start from the identity $(r+1)^3-r^3=3r^2+3r+1$.

Expand $(r+1)^3=r^3+3r^2+3r+1$ and subtract $r^3$.

Sum both sides from $r=1$ to $n$: $\displaystyle\sum_{r=1}^n\big[(r+1)^3-r^3\big]=\sum_{r=1}^n(3r^2+3r+1)$.

Summing an identity that holds for every $r$.

The left side telescopes: every term cancels except the first and last, leaving $(n+1)^3-1$.

$\sum[(r+1)^3-r^3]=(2^3-1^3)+(3^3-2^3)+\cdots+((n+1)^3-n^3)$; consecutive terms cancel.

$(n+1)^3-1=3\displaystyle\sum_{r=1}^nr^2+3\cdot\frac{n(n+1)}{2}+n$

Substituting Theorem 8.2 for $\sum r$ and $\sum 1=n$.

Expand and rearrange: $3\displaystyle\sum r^2=(n+1)^3-1-\frac{3n(n+1)}{2}-n$, which simplifies to $\dfrac{n(n+1)(2n+1)}{2}$.

Common-denominator algebra, factoring out $(n+1)$ throughout.

$\displaystyle\sum_{r=1}^nr^2=\dfrac{n(n+1)(2n+1)}{6}$

Divide both sides by 3.

∎

The sum of cubes

Theorem 9.2

$$\sum_{r=1}^{n}r^3=\left[\frac{n(n+1)}{2}\right]^2=\left(\sum_{r=1}^nr\right)^2$$ Remarkably, the sum of the first $n$ cubes equals the square of the sum of the first $n$ natural numbers.

Proof sketch — Theorem 9.2

This result can be proved by the same telescoping technique applied to $(r+1)^4-r^4=4r^3+6r^2+4r+1$, substituting the now-known formulae for $\sum r^2$ and $\sum r$, then simplifying — a longer but structurally identical argument to the proof of Theorem 9.1. It is also a standard target for proof by induction (see the Proof topic).

Example 1 · Direct evaluation of Σr²

Find $\displaystyle\sum_{r=1}^{15}r^2$.

$\dfrac{15(16)(31)}{6}=\dfrac{7440}{6}=1240$.

Example 2 · A sum not starting at r = 1

Find $\displaystyle\sum_{r=8}^{20}r^2$.

$\displaystyle\sum_{r=8}^{20}r^2=\sum_{r=1}^{20}r^2-\sum_{r=1}^{7}r^2=\frac{20(21)(41)}{6}-\frac{7(8)(15)}{6}=2870-140=2730$.

Example 3 · A combined quadratic series

Show that $\displaystyle\sum_{r=1}^{n}r(r-1)=\frac{n(n+1)(2n-2)}{6}$, simplifying to lowest terms.

$\displaystyle\sum_{r=1}^nr(r-1)=\sum_{r=1}^n(r^2-r)=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}$

$=\dfrac{n(n+1)}{6}\big[(2n+1)-3\big]=\dfrac{n(n+1)(2n-2)}{6}=\dfrac{n(n+1)(n-1)}{3}$.

Combining all three standard results

Method 9.3 — mixed linear, quadratic and cubic terms

Sigma notation stays linear no matter how many standard results are involved: a series like $\sum(ar^3+br^2+cr+d)$ splits term by term into separate multiples of $\sum r^3$, $\sum r^2$, $\sum r$ and $\sum 1$, each evaluated from its own formula and then recombined. The only new skill beyond Examples 1–3 is keeping track of three formulas at once instead of one or two — the algebra is otherwise identical.

Example 4 · All three formulas together

Show that $\displaystyle\sum_{r=1}^{n}(2r^3-r^2+3r)=\frac{n(n+1)(3n^2+n+8)}{6}$.

Split into three separate sums using linearity:

$\displaystyle\sum_{r=1}^n(2r^3-r^2+3r)=2\sum_{r=1}^nr^3-\sum_{r=1}^nr^2+3\sum_{r=1}^nr$

$=2\left[\frac{n(n+1)}{2}\right]^2-\frac{n(n+1)(2n+1)}{6}+3\cdot\frac{n(n+1)}{2}$

$=\dfrac{n^2(n+1)^2}{2}-\dfrac{n(n+1)(2n+1)}{6}+\dfrac{3n(n+1)}{2}$

Every term has a factor of $n(n+1)$, so take that out and put everything over a common denominator of 6:

$=\dfrac{n(n+1)}{6}\Big[3n(n+1)-(2n+1)+9\Big]=\dfrac{n(n+1)}{6}\Big[3n^2+3n-2n-1+9\Big]$

$=\dfrac{n(n+1)(3n^2+n+8)}{6}$, as required.

As a quick check, substitute $n=1$: the left side is $2(1)^3-1^2+3(1)=4$, and the right side gives $\dfrac{1(2)(3+1+8)}{6}=\dfrac{24}{6}=4$ ✓.

Example 5 · Proving a cube identity

Prove that $\displaystyle\sum_{r=1}^{n}(2r-1)^3=n^2(2n^2-1)$.

$(2r-1)^3=8r^3-12r^2+6r-1$.

$\displaystyle\sum_{r=1}^n(2r-1)^3=8\sum r^3-12\sum r^2+6\sum r-n$

$=8\left[\frac{n(n+1)}{2}\right]^2-12\cdot\frac{n(n+1)(2n+1)}{6}+6\cdot\frac{n(n+1)}{2}-n$

$=2n^2(n+1)^2-2n(n+1)(2n+1)+3n(n+1)-n$

Take out the common factor $n$: $=n\Big[2n(n+1)^2-2(n+1)(2n+1)+3(n+1)-1\Big]$

Expand each bracketed term separately: $2n(n+1)^2=2n(n^2+2n+1)=2n^3+4n^2+2n$; $\ 2(n+1)(2n+1)=2(2n^2+3n+1)=4n^2+6n+2$; $\ 3(n+1)=3n+3$.

So the bracket becomes $2n^3+4n^2+2n-4n^2-6n-2+3n+3-1=2n^3-n$.

Therefore $\displaystyle\sum_{r=1}^n(2r-1)^3=n(2n^3-n)=2n^4-n^2=n^2(2n^2-1)$, as required.

Exam technique

Always factor out the highest common factor (usually $n(n+1)$ or $n$) as early as possible when simplifying combinations of $\sum r$, $\sum r^2$, $\sum r^3$ — leaving everything over a common denominator of 6 until the final step keeps the algebra far more manageable than expanding early. When a target answer is given (a "show that" question), substituting a small value of $n$ into both your working and the target is a fast way to catch an arithmetic slip before it costs marks.

Exam-style questions

Question 1 6 marks

(a) State the formula for $\displaystyle\sum_{r=1}^nr^3$. (b) Find $\displaystyle\sum_{r=1}^{12}r^3$. (c) Hence find $\displaystyle\sum_{r=1}^{12}(r^3+r)$.

(a) $\left[\dfrac{n(n+1)}{2}\right]^2$ (B1)

(b) $\left[\dfrac{12(13)}{2}\right]^2=78^2=6084$ (M1, A1)

Question 2 7 marks

Show that $\displaystyle\sum_{r=1}^{n}(r^3-r^2)=\frac{n(n+1)(n-1)(3n+2)}{12}$.

$\displaystyle\sum(r^3-r^2)=\left[\frac{n(n+1)}{2}\right]^2-\frac{n(n+1)(2n+1)}{6}$ (M1, A1)

$=\dfrac{n(n+1)}{12}\big[3n(n+1)-2(2n+1)\big]=\dfrac{n(n+1)}{12}\big[3n^2+3n-4n-2\big]$ (M1, A1)

$=\dfrac{n(n+1)(3n^2-n-2)}{12}=\dfrac{n(n+1)(3n+2)(n-1)}{12}$ (M1, A1, A1)

Question 3 6 marks

Find $\displaystyle\sum_{r=10}^{25}r^3$.

$\displaystyle\sum_{r=1}^{25}r^3=\left[\frac{25(26)}{2}\right]^2=325^2=105625$ (M1, A1)

$\displaystyle\sum_{r=1}^{9}r^3=\left[\frac{9(10)}{2}\right]^2=45^2=2025$ (M1, A1)

$\displaystyle\sum_{r=10}^{25}r^3=105625-2025=103600$ (M1, A1)

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Sums of Natural Numbers

Roots of a Quadratic Equation