Maths › Further algebra and series › Summing series and the method of differences
Summing series and the method of differences
Two methods give a closed form for a long sum: the standard results for the sums of powers of r, and the method of differences, in which each term is written as f(r) − f(r + 1) so that intermediate terms cancel in pairs and only terms from each end remain.
Builds on Sequences and sigma notation and Partial fractions and Arithmetic series.
IN THIS TOPIC
- Quote and combine the standard results for Σr, Σr² and Σr³.
- Use Σr to count the comparisons an algorithm makes, read its order off the dominant term, and say what doubling the list does to the work.
- Split a term into a difference and telescope the sum.
COMMON MISCONCEPTION
There is no way to sum 1/(1×2) + 1/(2×3) + … + 1/(n(n+1)) exactly; you can only add the terms up.
Partial fractions give 1/(r(r + 1)) = 1/r − 1/(r + 1), so consecutive terms cancel in pairs and the sum is exactly 1 − 1/(n + 1). That is the method of differences: write each term as f(r) − f(r + 1), write out both ends, and state which terms remain. At n = 4 the formula gives 4/5, and 1/2 + 1/6 + 1/12 + 1/20 confirms it.
Standard results
Three closed forms cover most of this topic.
Check the current formula booklet for your awarding organisation before assuming that a standard sum is supplied; provision differs between organisations, and Σr is often not printed. Any polynomial in r can be summed by splitting it into these standard sums, since sigma distributes over sums and constant factors can be taken outside. Note that Σr³ is the square of Σr: the sum of the first n cubes is the square of the sum of the first n integers.
Σr is also what decision maths counts with. An algorithm that compares each item on a list with each of the others makes 1 + 2 + … + (n − 1) = ½n(n − 1) comparisons, which is ½n² − ½n. The order of an algorithm is read off the dominant term, the one that outgrows the rest as n rises, so the −½n and the constant factor are discarded: this method is order n², written O(n²). Doubling the length of the list therefore multiplies the number of comparisons by approximately four.
WORKED EXAMPLE
A polynomial series
Show that Σ r(r + 1) from 1 to n equals n(n + 1)(n + 2)/3.
Split the sum: Σ r(r + 1) = Σ r² + Σ r = n(n+1)(2n+1)/6 + n(n+1)/2.
Take out the common factor n(n + 1)/6 to get n(n+1)[(2n + 1) + 3]/6 = n(n+1)(2n+4)/6.
So the sum is n(n + 1)(n + 2)/3. Check at n = 4: 2 + 6 + 12 + 20 = 40, and 4 × 5 × 6/3 = 40.
The method of differences
The method of differences applies when each term can be written as f(r) − f(r + 1), usually by partial fractions. Write out the first few terms and the last few in full. Each intermediate value appears once with a plus sign and once with a minus sign, so the sum reduces to the terms at the two ends.
WORKED EXAMPLE
A classic telescope
Find Σ 1/(r(r + 1)) from 1 to n.
Partial fractions give 1/(r(r + 1)) = 1/r − 1/(r + 1).
The sum is (1 − 1/2) + (1/2 − 1/3) + … + (1/n − 1/(n + 1)).
All the inner terms cancel in pairs, so the total is 1 − 1/(n + 1).
Check at n = 4: 1/2 + 1/6 + 1/12 + 1/20 = 4/5, and 1 − 1/5 = 4/5.
GUIDED PRACTICE
A gap-two telescope
Given that 1/(r(r + 2)) = ½(1/r − 1/(r + 2)), find Σ 1/(r(r + 2)) from 1 to n.
Show the working
With a gap of two, each term cancels against the term two rows further on, so two terms remain at each end.
The sum is ½(1 + 1/2 − 1/(n + 1) − 1/(n + 2)).
Tidied, that is 3/4 − (2n + 3)/(2(n + 1)(n + 2)). Check at n = 3: 1/3 + 1/8 + 1/15 = 21/40, and ½(3/2 − 1/4 − 1/5) = 21/40.
ASSESSMENT FOCUS
- Check the current formula booklet for your awarding organisation; Σr = ½n(n + 1) is often not supplied.
- The standard results start at r = 1. For a sum from r = k, subtract the sum up to k − 1.
- The order is read from the dominant term, with lower powers and constant factors discarded, so a count of ½n² − ½n is O(n²).
- Factorise while combining standard results rather than expanding first.
- In a method-of-differences answer, write at least two terms at each end before cancelling, and state which terms remain.
- A gap of two in the denominators leaves two terms at each end rather than one.
- When a question says 'hence', use the split obtained in the previous part.
CHECK YOURSELF
Evaluate Σ r² for r from 1 to 10.
Show a hint
n(n + 1)(2n + 1)/6 with n = 10.
Show the answer
10 × 11 × 21/6 = 385. Adding the ten squares directly confirms it, since 1 + 4 + 9 + … + 100 = 385.
Polynomial series split into Σr³, Σr² and Σr; check which of these your awarding organisation's formula booklet supplies.
The method of differences writes each term as f(r) − f(r + 1); write out both ends and state the terms that remain.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the summing series and the method of differences questions page.
USED LATER IN
CHECK YOUR PROGRESS
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- Quote and combine the standard results for Σr, Σr² and Σr³.
- Use Σr to count the comparisons an algorithm makes, read its order off the dominant term, and say what doubling the list does to the work.
- Split a term into a difference and telescope the sum.
Open the full revision checklist to see every objective in the course in one place.