Maths › Further algebra and series › Summing series and the method of differences
Summing series and the method of differences
Two ways to reduce a long sum to a short formula. Standard results for the sums of powers of r, and the method of differences, in which intermediate terms cancel in pairs and only terms from the beginning and the end remain.
Builds on Sequences and sigma notation and Partial fractions.
IN THIS TOPIC
- Quote and combine the standard results for Σr, Σr² and Σr³.
- 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.
Standard results
Three closed forms cover most of this topic, and they do not all come from the same place. The first one is not in the booklet. Learn it.
The other two are printed, under Summations, so look them up rather than trusting your memory of which is which.
That split catches people out. Candidates open the booklet expecting the whole trio, find Σr missing, and stall. Any polynomial in r sums by splitting into these pieces, since sigma distributes over sums and constants pull out. Notice that Σr³ is the square of Σr, so the sum of the first n cubes is the square of their sum.
WORKED EXAMPLE
A polynomial series
Show that Σ r(r + 1) from 1 to n equals n(n + 1)(n + 2)/3.
Split it up. 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 telescope
The method of differences applies when each term splits as f(r) − f(r + 1), usually via partial fractions. Write the first few rows and the last few in full. Everything in the middle appears once with a plus and once with a minus, so the sum collapses 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, terms cancel two rows down, so two terms survive 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
- Σr² and Σr³ are printed in the booklet under Summations. Σr = ½n(n + 1) is not, so carry that one in your head.
- Standard results start at r = 1. For a sum from r = k, subtract the sum up to k − 1.
- Factorise early when combining standard results; expanding everything first buries the answer.
- In a telescope, write at least two rows at each end before cancelling, and say what survives.
- A gap of two in the denominators leaves two survivors at each end, not one.
- If a question says 'hence', the previous part's split is the intended tool. Do not start again.
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, Σr. Two of those are in the booklet and Σr is not.
Telescopes split each term into a difference. Write both ends and keep the survivors.
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.
CHECK YOUR PROGRESS
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- Quote and combine the standard results for Σr, Σr² and Σr³.
- 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.