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Sums of seriesEdexcel A-Level Further Maths: Revision notes

Section 1

The standard results

Three results let you sum any series whose general term is a polynomial in rr: ∑r=1nr=12n(n+1),∑r=1nr2=16n(n+1)(2n+1),∑r=1nr3=14n2(n+1)2.\sum_{r=1}^{n}r=\frac12n(n+1),\qquad\sum_{r=1}^{n}r^2=\frac16n(n+1)(2n+1),\qquad\sum_{r=1}^{n}r^3=\frac14n^2(n+1)^2. A constant term sums to nn times the constant: ∑r=1nc=cn\sum_{r=1}^{n}c=cn. These are in the formulae booklet, but you must be fluent in using them. Note that ∑r3=(∑r)2\sum r^3=\left(\sum r\right)^2.

Key termsseriessum of squaressum of cubes
Common mistake

Treating ∑r=1n1\sum_{r=1}^{n}1 as 11. It is nn, because you are adding nn ones.

Exam tip

Check any closed form by substituting n=1n=1 and n=2n=2 and comparing with the terms added directly.

Section 2

Summing other series

Expand the general term into powers of rr, split the sum, take out constants, and use the standard results. Example: ∑r=1nr(r2+2)=∑r3+2∑r=14n2(n+1)2+n(n+1)=14n(n+1)[n(n+1)+4]=14n(n+1)(n2+n+4)\sum_{r=1}^{n}r(r^2+2)=\sum r^3+2\sum r=\frac14n^2(n+1)^2+n(n+1)=\frac14n(n+1)\left[n(n+1)+4\right]=\frac14n(n+1)(n^2+n+4). Example: ∑r=1n(2r−1)2=4∑r2−4∑r+n=13n(4n2−1)\sum_{r=1}^{n}(2r-1)^2=4\sum r^2-4\sum r+n=\frac13n(4n^2-1). Always factorise the answer: take out the common factor of nn (and usually (n+1)(n+1)) before simplifying the bracket. This makes 'show that' questions straightforward and gives a neater result.

Key termsfactorise
Common mistake

Writing ∑r2=(∑r)2\sum r^2=\left(\sum r\right)^2. That is true for cubes, not squares.

Common mistake

Failing to expand (2r−1)2(2r-1)^2 before summing: there is no formula for a sum of a bracket squared.

Section 3

Sums that do not start at r=1r=1

The formulae only work from r=1r=1. For a sum from r=ar=a to r=br=b use the difference of two sums from 1: ∑r=abur=∑r=1bur−∑r=1a−1ur.\sum_{r=a}^{b}u_r=\sum_{r=1}^{b}u_r-\sum_{r=1}^{a-1}u_r. Example: ∑r=1020r(r2+2)=f(20)−f(9)=44520−2115=42405\sum_{r=10}^{20}r(r^2+2)=f(20)-f(9)=44520-2115=42405, where f(n)=14n(n+1)(n2+n+4)f(n)=\frac14n(n+1)(n^2+n+4). The same idea sums a list written out in words: 512+532+⋯+992=∑r=2650(2r−1)251^2+53^2+\cdots+99^2=\sum_{r=26}^{50}(2r-1)^2.

Key termslimits
Common mistake

Subtracting f(a)f(a) instead of f(a−1)f(a-1). This loses the first term of the range.

Exam tip

To turn a list into a sum, find the values of rr for the first and last terms by solving 2r−1=512r-1=51 and 2r−1=992r-1=99 (or similar).

Section 4

Using a closed form

Once you have ∑r=1nur\sum_{r=1}^{n}u_r in terms of nn, you can substitute a value of nn or use it in later parts. A closed form that is already known can also be reused: since (2r−1)(2r)=(2r−1)2+(2r−1)(2r-1)(2r)=(2r-1)^2+(2r-1) and ∑r=1n(2r−1)=n2\sum_{r=1}^{n}(2r-1)=n^2, we get ∑r=1n(2r−1)(2r)=13n(4n2−1)+n2=13n(n+1)(4n−1)\sum_{r=1}^{n}(2r-1)(2r)=\frac13n(4n^2-1)+n^2=\frac13n(n+1)(4n-1). For a series such as ∑r=1n(r2+r)=13n(n+1)(n+2)\sum_{r=1}^{n}\left(r^2+r\right)=\frac13n(n+1)(n+2), test with n=6n=6: 91+21=11291+21=112 and 6×7×83=112\frac{6\times7\times8}{3}=112.

Key termsclosed form
Exam tip

In a 'hence' question, rewrite the new series in terms of the one you have just summed rather than starting again.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Sums of series

  1. A student uses the standard summation formulae for the first nn positive integers, their squares and their cubes.
    Find the value of ∑r=112r3\sum_{r=1}^{12}r^3.2 marks
  2. Let Sn=∑r=1n(r2+r)S_n=\sum_{r=1}^{n}\left(r^2+r\right).
    Hence find the value of ∑r=1120(r2+r)\sum_{r=11}^{20}\left(r^2+r\right).2 marks
  3. Let f(n)=∑r=1nr(r2+2)f(n)=\sum_{r=1}^{n}r\left(r^2+2\right).
    Show that f(n)=14n(n+1)(n2+n+4)f(n)=\frac14n(n+1)\left(n^2+n+4\right).3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).