1.3 - Sums of powers | Cambridge A-Level Further Mathematics

A long sum can have a short exact answer. You will use the three standard power sums in 1.3 to add polynomial terms, then adapt the method when the starting point or the index changes.

The three standard sums

How could you find $1^3+2^3+\cdots+100^3$ without entering a hundred terms? Its value is the square of a much simpler sum.

First, $1+2+3+4+5=15$ . Pairing the first and last terms gives $1+5=2+4=6$ , with the middle $3$ left over. More generally, writing $1+2+\cdots+n$ once forwards and once backwards gives $n$ pairs, each equal to $n+1$ . Dividing by $2$ gives $n(n+1)/2$ .

Sigma notation

$\displaystyle\sum_{r=a}^{b}f(r)$ means add $f(a),f(a+1),\ldots,f(b)$ , with integer steps and both endpoints included. The letter $r$ is a summation index: it changes within the sum; $a$ and $b$ specify its range.

For example, $\sum_{r=1}^{4}r^2=1+4+9+16=30$ . The square applies to each term before addition. It does not mean $(1+2+3+4)^2=100$ .

Standard power sums

For a positive integer $n$ :

\begin{aligned} \sum_{r=1}^{n}r&=\frac{n(n+1)}2 \\ \sum_{r=1}^{n}r^2&=\frac{n(n+1)(2n+1)}6 \\ \sum_{r=1}^{n}r^3&=\frac{n^2(n+1)^2}4=\left(\frac{n(n+1)}2\right)^2. \end{aligned}

In the formula booklet.

The last identity says that the sum of the cubes is the square of the sum of the integers. For $n=3$ , this is $1+8+27=(1+2+3)^2=36$ . These are three different summands, so choose the formula by inspecting the term being added, not the shape of the final answer.

Worked example 1: adding squares

Question. Evaluate $\displaystyle\sum_{r=1}^{10}r^2$ .

Plan. Use the square-sum formula with upper limit $n=10$ .

\sum_{r=1}^{10}r^2 =\frac{10(11)(21)}6 =385.

The lower limit is $1$ , so the formula applies directly.

Answer. The sum is $385$ .

Check. Direct addition gives $1+4+9+16+25+36+49+64+81+100=385$ .

Keep the same limits, but change the power from $2$ to $3$ . This changes which standard result you need.

Worked example 2: adding cubes

Question. Evaluate $\displaystyle\sum_{r=1}^{10}r^3$ .

Plan. Find the sum of the integers, then square it using the cube-sum identity.

\sum_{r=1}^{10}r^3 =\left(\frac{10(11)}2\right)^2 =55^2 =3025.

Answer. The sum is $3025$ .

Check. Adding the cubes directly gives $1+8+27+64+125+216+343+512+729+1000=3025$ .

The cube-sum identity is special: do not extend it to other powers without proof.

Watch out

$\sum r^2=(\sum r)^2$ is false. For $r=1,2$ , the left side is $5$ , whereas the right side is $9$ . The standard square-sum and cube-sum formulae are different.

Use the formula for the actual summand in this check.

Related polynomial sums

A polynomial summand can be separated into the standard powers by expanding it before summing.

For instance, $(r+1)(r+2)=r^2+3r+2$ . Adding this expression for every $r$ means adding all the squares, all the $3r$ terms and all the constants separately:

\sum_{r=1}^{n}(r^2+3r+2) =\sum_{r=1}^{n}r^2+3\sum_{r=1}^{n}r+2n.

The last term is $2n$ , because there are $n$ copies of $2$ . A constant does not disappear from a sum.

You already know

Expansion reveals the powers present. Summation distributes over addition and subtraction, and a constant factor can be taken outside the sum.

Keep common factors visible when you substitute the formulae. It is usually easier to factor a short bracket than to expand everything into a quartic and factor it again.

Worked example 3: a quadratic summand

Question. Find $\displaystyle\sum_{r=1}^{n}(r+1)(r+2)$ in terms of $n$ .

Plan. Expand, use the standard sums, and then collect the factors.

\begin{aligned} \sum_{r=1}^{n}(r+1)(r+2) &=\sum_{r=1}^{n}(r^2+3r+2) \\ &=\frac{n(n+1)(2n+1)}6+\frac{3n(n+1)}2+2n \\ &=\frac n6\bigl((n+1)(2n+1)+9(n+1)+12\bigr) \\ &=\frac n6(2n^2+12n+22). \end{aligned}

The constant contribution is included as $2n$ before simplifying.

Answer. $\displaystyle\frac{n(n^2+6n+11)}3$ .

Check. For $n=2$ , the original terms are $2\times3+3\times4=18$ . The formula gives $2(4+12+11)/3=18$ .

Now multiply the summand by $r$ . This raises its degree from $2$ to $3$ , so the cube-sum formula joins the calculation.

Worked example 4: a cubic summand

Question. Find $\displaystyle\sum_{r=1}^{n}r(r+1)(r+2)$ in factorised form.

Plan. Expand into three standard powers, then factor out $n(n+1)$ .

\begin{aligned} \sum_{r=1}^{n}r(r+1)(r+2) &=\sum_{r=1}^{n}(r^3+3r^2+2r) \\ &=\frac{n^2(n+1)^2}4 +\frac{n(n+1)(2n+1)}2+n(n+1) \\ &=\frac{n(n+1)}4\bigl(n(n+1)+2(2n+1)+4\bigr) \\ &=\frac{n(n+1)}4(n^2+5n+6). \end{aligned}

Since $n^2+5n+6=(n+2)(n+3)$ , all four consecutive factors appear.

Answer. $\displaystyle\frac{n(n+1)(n+2)(n+3)}4$ .

Check. For $n=3$ , direct addition gives $6+24+60=90$ . The expression gives $3(4)(5)(6)/4=90$ .

The two examples use the same expansion, but the extra factor of $r$ changes every power. Re-expand the whole summand before choosing formulae.

Problem-solving

Expand the summand first; substitute standard sums second; simplify last. Keep factors common to all terms outside a bracket.

Try a shifted pair of brackets, retaining the outer factor of $r$ .

Changing the limits

The standard formulae start at $1$ , so a different starting point requires you to remove an initial section.

Write $F(n)=\sum_{r=1}^{n}r^2=n(n+1)(2n+1)/6$ , with $F(0)=0$ . Then $\sum_{r=4}^{9}r^2=F(9)-F(3)$ : remove the terms ending at $3$ , so the term at $4$ remains.

More generally, for integers $1\leq a\leq b$ ,

\sum_{r=a}^{b}f(r) =\sum_{r=1}^{b}f(r)-\sum_{r=1}^{a-1}f(r).

There are $b-a+1$ terms, not $b-a$ . This is also a useful check when a constant appears in the summand.

Worked example 5: limits that depend on n

Question. Show that, for a positive integer $n$ ,

\sum_{r=n+1}^{2n}r^2=\frac{n(2n+1)(7n+1)}6.

Plan. Subtract the first $n$ squares from the first $2n$ squares.

\begin{aligned} \sum_{r=n+1}^{2n}r^2 &=\frac{2n(2n+1)(4n+1)}6-\frac{n(n+1)(2n+1)}6 \\ &=\frac{n(2n+1)}6\bigl(2(4n+1)-(n+1)\bigr) \\ &=\frac{n(2n+1)(7n+1)}6. \end{aligned}

The factor $n(2n+1)$ is common to both terms, so keep it outside the bracket.

Answer. The required identity holds for every positive integer $n$ .

Check. At $n=3$ , the original sum is $4^2+5^2+6^2=77$ ; the result gives $3(7)(22)/6=77$ .

Change only the term from $r^2$ to $(r+1)^2$ . The squares being added now start and finish one place later; both boundaries must move.

Worked example 6: shifting the index

Question. Find $\displaystyle\sum_{r=n+1}^{2n}(r+1)^2$ in factorised form.

Plan. Set $s=r+1$ , change both limits, then subtract standard sums.

When $r=n+1$ , $s=n+2$ ; when $r=2n$ , $s=2n+1$ . Hence

\begin{aligned} \sum_{r=n+1}^{2n}(r+1)^2 &=\sum_{s=n+2}^{2n+1}s^2 \\ &=\frac{(2n+1)(2n+2)(4n+3)}6 -\frac{(n+1)(n+2)(2n+3)}6 \\ &=\frac{n+1}{6}\bigl(2(2n+1)(4n+3)-(n+2)(2n+3)\bigr) \\ &=\frac{n+1}{6}(14n^2+13n). \end{aligned}

Answer. $\displaystyle\frac{n(n+1)(14n+13)}6$ .

Check. Compared with Worked example 5, the shifted sum gains $(2n+1)^2$ and loses $(n+1)^2$ . Their difference is $3n^2+2n$ ; subtracting the two final formulae gives exactly this difference.

You could instead expand $(r+1)^2$ and subtract initial sections for each power. Changing the index often keeps the algebra shorter.

Watch out

To sum from $a$ , subtract the initial sum ending at $a-1$ . Subtracting through $a$ removes the first term you wanted to keep.

Apply both endpoint ideas in the following sum.

Quick Recap

Identify whether the summand involves $r$ , $r^2$ , $r^3$ or a combination.
Expand polynomial terms and sum each power separately; a constant appears once per term.
Subtract the initial section ending one place before the required lower limit.
When changing the index, change both endpoints.

The opening sum is $\bigl(100(101)/2\bigr)^2=25\,502\,500$ : a hundred cubes reduce to one square.

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1.3 - Sums of powers

The three standard sums

Related polynomial sums

Changing the limits

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