MathsFurther 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.

Σr=n(n+1)2Σ r = \frac{n(n+1)}{2}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.

Σr2=n(n+1)(2n+1)6,Σr3=n2(n+1)24Σ r^{2} = \frac{n(n+1)(2n+1)}{6}, Σ r^{3} = \frac{n^{2}(n+1)^{2}}{4}IN THE FORMULAE BOOKLET

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.

The first five squares as columns: their total 55 is n(n + 1)(2n + 1)/6 with n = 51r = 14r = 29r = 316r = 425r = 51 + 4 + 9 + 16 + 25 = 55
FIG. 1The first five squares stacked as columns: 1 + 4 + 9 + 16 + 25 = 55, exactly n(n + 1)(2n + 1)/6 with n = 5.

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.

The method of differences on 1/(r(r + 1)): each row's right term cancels the next row's left term1− 1/21/2− 1/31/3− 1/41/n− 1/(n + 1)everything betweencancels in pairssum = 1 − 1/(n + 1)
FIG. 2The telescoping sum for 1/(r(r + 1)): every middle fraction appears twice and cancels, leaving only 1 and the final −1/(n + 1).

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.

6 questions on this topicAnswer them one at a time and mark yourself against the worked answer.Practise this topic

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.