A-Level Further Maths 9231 / Topic 1.3 / AS & A Level

Add a hundred terms in one line.

Three formulas from the formula list let you add up \(r\), \(r^{2}\) and \(r^{3}\) for \(r = 1\) to \(n\) without writing out a single term. Combine them with two simple rules and you can sum any polynomial in \(r\), start the sum anywhere, skip every other term, or work backwards from a total to find \(n\).

Topic 1.3All summation of series lessons →

The key idea

\(\displaystyle\sum_{r=1}^{n} r = \tfrac{1}{2}n(n+1)\), \(\displaystyle\sum_{r=1}^{n} r^{2} = \tfrac{1}{6}n(n+1)(2n+1)\), \(\displaystyle\sum_{r=1}^{n} r^{3} = \tfrac{1}{4}n^{2}(n+1)^{2}\). Sums split term by term, constants come out in front, and \(\displaystyle\sum_{r=1}^{n} c = cn\). A sum from \(r = m\) to \(n\) is the sum to \(n\) minus the sum to \(m - 1\).

Fig. 1: two copies of the staircase 1 + 2 + 3 + 4 + 5 fit together into a 5 by 6 rectangle, so the sum is half of 5 × 6. The same picture with n rows gives ½n(n + 1)

Section 01

Sigma notation and the three results.

\(\displaystyle\sum_{r=1}^{n} f(r)\) means: put \(r = 1, 2, 3, \dots, n\) into \(f(r)\) and add the results. The letter \(r\) is only a counter; the answer depends on \(n\), not on \(r\). The three standard results are given in the formula list, but you should know them by heart.

Sum	Standard result	Check with \(n = 3\)
\(\displaystyle\sum_{r=1}^{n} r\)	\(\tfrac{1}{2}n(n+1)\)	\(1 + 2 + 3 = 6\)
\(\displaystyle\sum_{r=1}^{n} r^{2}\)	\(\tfrac{1}{6}n(n+1)(2n+1)\)	\(1 + 4 + 9 = 14\)
\(\displaystyle\sum_{r=1}^{n} r^{3}\)	\(\tfrac{1}{4}n^{2}(n+1)^{2}\)	\(1 + 8 + 27 = 36\)

Split: \(\sum (f(r) + g(r)) = \sum f(r) + \sum g(r)\).
Constants out: \(\sum k f(r) = k \sum f(r)\).
A constant term: \(\displaystyle\sum_{r=1}^{n} c = c + c + \dots + c = cn\), since there are \(n\) terms.
Notice that \(\sum r^{3} = \left(\sum r\right)^{2}\). The results can be proved by induction (section 1.7) or by the method of differences (next lesson).

Worked example A linear sum

Evaluate \(\displaystyle\sum_{r=1}^{20} (3r + 2)\).

Split and take out the constant: \(3\displaystyle\sum_{r=1}^{20} r + \sum_{r=1}^{20} 2\).
\(\displaystyle\sum_{r=1}^{20} r = \tfrac{1}{2}(20)(21) = 210\).
\(\displaystyle\sum_{r=1}^{20} 2 = 2 \times 20 = 40\) (twenty terms, each 2).
Total: \(3(210) + 40\).

Answer: \(670\)

Worked example Cubes minus squares

Evaluate \(\displaystyle\sum_{r=1}^{10} (r^{3} - r^{2})\).

\(\displaystyle\sum_{r=1}^{10} r^{3} = \tfrac{1}{4}(10^{2})(11^{2}) = \tfrac{1}{4}(100)(121) = 3025\).
\(\displaystyle\sum_{r=1}^{10} r^{2} = \tfrac{1}{6}(10)(11)(21) = 385\).
Subtract: \(3025 - 385\).

Answer: \(2640\)

Your turn

Evaluate \(\displaystyle\sum_{r=1}^{15} (r^{2} + 1)\).

Examiner trap

\(\displaystyle\sum_{r=1}^{n} 5\) is \(5n\), not \(5\): the constant is added once for every value of \(r\). Forgetting this is the commonest slip when a polynomial in \(r\) has a constant term.

Section 02

Polynomials in r, fully factorised.

To sum a polynomial in \(r\), expand it into powers of \(r\), apply the three results, then simplify. Exam questions usually ask for the answer 'in a fully factorised form', so take out common factors such as \(n(n+1)\) BEFORE you expand anything.

Expand the general term: \(r(r+2) = r^{2} + 2r\), \((2r-1)^{2} = 4r^{2} - 4r + 1\).
Substitute the standard results, keeping them in factorised form.
Factor out everything common (often \(\tfrac{1}{6}n(n+1)\) or \(\tfrac{1}{3}n\)), simplify the bracket that is left, and factorise it if you can.
Check with \(n = 1\) or \(n = 2\): the formula must give the first term, then the sum of the first two.

Worked example Sum of r(r + 2)

Find \(\displaystyle\sum_{r=1}^{n} r(r + 2)\), giving your answer in a fully factorised form.

\(\displaystyle\sum_{r=1}^{n} r(r+2) = \sum r^{2} + 2\sum r = \tfrac{1}{6}n(n+1)(2n+1) + n(n+1)\).
Take out \(\tfrac{1}{6}n(n+1)\): \(\tfrac{1}{6}n(n+1)\left[(2n+1) + 6\right]\).
Simplify the bracket: \(2n + 7\).
Check \(n = 2\): \(1(3) + 2(4) = 11\), and \(\tfrac{1}{6}(2)(3)(11) = 11\).

Answer: \(\tfrac{1}{6}n(n+1)(2n+7)\)

Worked example Sum of (2r - 1) squared

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

\((2r - 1)^{2} = 4r^{2} - 4r + 1\), so the sum is \(4\sum r^{2} - 4\sum r + \sum 1\).
\(= \tfrac{2}{3}n(n+1)(2n+1) - 2n(n+1) + n\).
Take out \(\tfrac{1}{3}n\): \(\tfrac{1}{3}n\left[2(n+1)(2n+1) - 6(n+1) + 3\right] = \tfrac{1}{3}n\left[4n^{2} + 6n + 2 - 6n - 6 + 3\right]\).
The bracket is \(4n^{2} - 1 = (2n-1)(2n+1)\), a difference of two squares.

Answer: \(\tfrac{1}{3}n(2n-1)(2n+1)\), as required (check \(n = 2\): \(1 + 9 = 10 = \tfrac{1}{3}(2)(3)(5)\))

Your turn

Find \(\displaystyle\sum_{r=1}^{n} r(r + 1)\) in a fully factorised form.

Examiner trap

Do not expand everything into one cubic and then try to factorise it from scratch. Taking out \(n(n+1)\) at the start keeps the work short, and a factor of 2 hiding in the last bracket (as in \(2n + 4\)) must be taken out too, or the answer is not fully factorised.

Stage 1 · Learn

Check your understanding

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Evaluate \(\displaystyle\sum_{r=1}^{10} r^{2}\).

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What is \(\displaystyle\sum_{r=1}^{n} 4\)?

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

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Evaluate \(\displaystyle\sum_{r=11}^{20} r\).

Section 03

Other starting points, odd terms, and finding n.

The standard results all start at \(r = 1\). For any other sum, rewrite it in terms of sums that do start at 1. Write \(S(n)\) for the sum from \(r = 1\) to \(n\).

A later start: \(\displaystyle\sum_{r=m}^{n} f(r) = S(n) - S(m - 1)\). For example, \(\displaystyle\sum_{r=n+1}^{2n} f(r) = S(2n) - S(n)\): replace \(n\) by \(2n\) in the formula, then subtract.
Even terms only: \(2^{2} + 4^{2} + \dots + (2n)^{2} = \sum (2r)^{2} = 4\sum r^{2}\).
Odd terms only: either sum the general odd term \(2r - 1\), or take (all terms up to \(2n\)) minus (the even terms).
Finding \(n\): set the formula equal to the given total and solve; \(n\) must be a positive integer, so reject any other root.

Worked example From n + 1 to 2n

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

\(S(2n) = \tfrac{1}{6}(2n)(2n+1)(4n+1)\), replacing every \(n\) in \(\tfrac{1}{6}n(n+1)(2n+1)\) by \(2n\).
The sum is \(S(2n) - S(n) = \tfrac{1}{6}n(2n+1)\left[2(4n+1) - (n+1)\right]\), taking out \(\tfrac{1}{6}n(2n+1)\).
The bracket is \(8n + 2 - n - 1 = 7n + 1\).
Check \(n = 1\): the sum is just \(2^{2} = 4\), and \(\tfrac{1}{6}(1)(3)(8) = 4\).

Answer: \(\tfrac{1}{6}n(2n+1)(7n+1)\); for example \(21^{2} + 22^{2} + \dots + 40^{2} = \tfrac{1}{6}(20)(41)(141) = 19270\)

Worked example Work back to n

Given that \(\displaystyle\sum_{r=1}^{n} r^{3} = 44100\), find \(n\).

\(\tfrac{1}{4}n^{2}(n+1)^{2} = \left[\tfrac{1}{2}n(n+1)\right]^{2} = 44100\).
Both sides are positive squares, so \(\tfrac{1}{2}n(n+1) = 210\), giving \(n^{2} + n - 420 = 0\).
\((n - 20)(n + 21) = 0\); \(n\) is a positive integer, so reject \(n = -21\).

Answer: \(n = 20\)

Your turn

Show that the sum of the first \(n\) odd numbers, \(1 + 3 + 5 + \dots + (2n - 1)\), is \(n^{2}\).

Examiner trap

\(\displaystyle\sum_{r=11}^{20} r\) is \(S(20) - S(10)\), not \(S(20) - S(11)\): subtracting \(S(11)\) removes the term \(r = 11\), which belongs in the sum. Subtract the sum up to ONE LESS than the starting value.

Stage 2 · Exam

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Find \(\displaystyle\sum_{r=1}^{n} r(3r - 1)\), giving your answer in a fully factorised form.

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

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Given that \(\displaystyle\sum_{r=1}^{n} (2r + 3) = 320\), find the value of \(n\).

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Find \(2^{2} + 4^{2} + 6^{2} + \dots + (2n)^{2}\).

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Series summed without writing them out.

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Six original Cambridge-style questions

Evaluating standard sums, polynomials in \(r\) in fully factorised form, a product of three consecutive integers, a sum from \(n + 1\) to \(2n\), the sum of odd cubes and the least \(n\) for a sum to pass a target.