Maths › Further algebra and series › Summing series and the method of differences
Summing series and the method of differences
Two ways to collapse a long sum into a short formula: standard results for powers of r, and a telescope that cancels almost every term in sight.
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 f(r) − f(r + 1) form and telescope the sum.
- Handle sums that start above r = 1 by subtracting a shorter sum.
WHAT YOU PROBABLY THINK
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 carry the whole topic:
Any polynomial in r sums by splitting into these pieces, since sigma distributes over sums and constants pull out. Notice the third result is the square of the first: the sum of the first n cubes is the square of the first n integers' sum.
WORKED EXAMPLE
A polynomial series
Show that Σ r(r + 1) from 1 to n equals n(n + 1)(n + 2)/3.
Split: Σ r(r + 1) = Σ r² + Σ r = n(n+1)(2n+1)/6 + n(n+1)/2.
Common factor n(n + 1)/6: this is 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, and the sum collapses to the surviving ends. That collapse is what defeats the opening claim.
WORKED EXAMPLE
A classic telescope
Find Σ 1/(r(r + 1)) from 1 to n.
Partial fractions: 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: 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.
YOUR TURN
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: 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.
THE EXAM BIT
- Standard results start at r = 1; for a sum from r = k, subtract the sum 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.
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
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Polynomial series: split into Σr³, Σr², Σr, quote the closed forms, factorise early.
Telescopes: split each term as f(r) − f(r + 1), write both ends, 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.
CHECK YOUR PROGRESS
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- Quote and combine the standard results for Σr, Σr² and Σr³.
- Split a term into f(r) − f(r + 1) form and telescope the sum.
- Handle sums that start above r = 1 by subtracting a shorter sum.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.