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Summation of finite seriesEdexcel International A Level Further Maths: Revision notes

Section 1

Sigma notation

∑r=1nf(r)\sum_{r=1}^{n}f(r) means f(1)+f(2)+⋯+f(n)f(1)+f(2)+\dots+f(n): the sum of f(r)f(r) as rr runs from 11 to nn. Two facts follow directly. The sum of a constant is the constant multiplied by the number of terms, ∑r=1nc=cn\sum_{r=1}^{n}c=cn (in particular ∑r=1n1=n\sum_{r=1}^{n}1=n), and a sum can be split term by term and constants taken outside: ∑r=1n[af(r)+bg(r)]=a∑r=1nf(r)+b∑r=1ng(r).\sum_{r=1}^{n}\left[af(r)+bg(r)\right]=a\sum_{r=1}^{n}f(r)+b\sum_{r=1}^{n}g(r). The expression ∑r=1nr(r+2)\sum_{r=1}^{n}r(r+2) stands for 1×3+2×4+⋯+n(n+2)1\times3+2\times4+\dots+n(n+2); the answer is a function of nn.

Key termssigma notationfinite seriesconstant term
Common mistake

Writing ∑r=1n3=3\sum_{r=1}^{n}3=3. There are nn terms, each equal to 33, so the sum is 3n3n.

Section 2

The standard results

You must know the sum of the first nn positive integers: ∑r=1nr=n(n+1)2.\sum_{r=1}^{n}r=\frac{n(n+1)}{2}. It comes from writing the sum forwards and backwards: each pair of terms adds to n+1n+1 and there are nn pairs, giving n(n+1)n(n+1) for twice the sum. The result for squares is in the formulae booklet: ∑r=1nr2=n(n+1)(2n+1)6.\sum_{r=1}^{n}r^2=\frac{n(n+1)(2n+1)}{6}. Check: for n=3n=3, 1+4+9=14=3×4×761+4+9=14=\frac{3\times4\times7}{6}. The method of differences is not required.

Key termsstandard resultsum of integerssum of squares
Exam tip

Test any formula you quote on n=1n=1 and n=2n=2 before using it.

Section 3

Summing expressions such as r(r+2)r(r+2)

To sum a quadratic in rr, expand it, split it and apply the standard results: ∑r=1nr(r+2)=∑r2+2∑r=n(n+1)(2n+1)6+n(n+1).\sum_{r=1}^{n}r(r+2)=\sum r^2+2\sum r=\frac{n(n+1)(2n+1)}{6}+n(n+1). Take out the common factor n(n+1)6\frac{n(n+1)}{6}: n(n+1)6[(2n+1)+6]=n(n+1)(2n+7)6\frac{n(n+1)}{6}\left[(2n+1)+6\right]=\frac{n(n+1)(2n+7)}{6}. Check: n=10n=10 gives 10×11×276=495\frac{10\times11\times27}{6}=495, and ∑r2+2∑r=385+110=495\sum r^2+2\sum r=385+110=495. If the expression contains constants, use ∑c=cn\sum c=cn, for example ∑r=1n(2r+3)=n(n+1)+3n=n(n+4)\sum_{r=1}^{n}(2r+3)=n(n+1)+3n=n(n+4).

Key termsexpandcommon factor
Common mistake

Using ∑r2=(∑r)2\sum r^2=\left(\sum r\right)^2. The sum of squares is not the square of the sum.

Section 4

Factorising and 'show that' answers

Examiners usually want a fully factorised form. When both standard results appear, factorise n(n+1)6\frac{n(n+1)}{6} first, then simplify the bracket. For a 'show that', write each stage: split the sum, substitute the standard results, take out the common factor, and arrive at the given form. Example: ∑r=1nr(2r+3)=2⋅n(n+1)(2n+1)6+3⋅n(n+1)2=n(n+1)6[2(2n+1)+9]=n(n+1)(4n+11)6\sum_{r=1}^{n}r(2r+3)=2\cdot\frac{n(n+1)(2n+1)}{6}+3\cdot\frac{n(n+1)}{2}=\frac{n(n+1)}{6}[2(2n+1)+9]=\frac{n(n+1)(4n+11)}{6}.

Key termsshow thatfactorise
Exam tip

Do not expand n(n+1)(2n+1)n(n+1)(2n+1) unless you have to; keeping factors makes the next step easier.

Section 5

Sums between limits and solving for nn

To sum from r=ar=a to r=br=b, subtract: ∑r=abf(r)=∑r=1bf(r)−∑r=1a−1f(r)\sum_{r=a}^{b}f(r)=\sum_{r=1}^{b}f(r)-\sum_{r=1}^{a-1}f(r). Example: the 6th to the 10th terms of r(r+2)r(r+2) sum to S10−S5=495−85=410S_{10}-S_5=495-85=410. To find nn given SnS_n equal to a number, set up the equation, expand to a polynomial, find one root by substitution and factorise; use the discriminant to show that any quadratic factor has no real roots. Remember nn is a positive integer.

Key termslower limitupper limit
Common mistake

Subtracting SaS_a instead of Sa−1S_{a-1} when summing from r=ar=a; the aath term would be lost.

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Exam questions on Summation of finite series

  1. A sequence has rrth term ur=2r+3u_r=2r+3, and SnS_n is the sum of the first nn terms.
    Find the smallest value of nn for which Sn>1000S_n>1000.2 marks
  2. Let Sn=1×3+2×4+3×5+⋯+n(n+2)S_n=1\times3+2\times4+3\times5+\dots+n(n+2), the sum of the first nn terms of a series.
    Find the sum of the 6th to the 10th terms of the series, inclusive.2 marks
  3. An orange display is built in layers. Layer rr, counting from the top with r=1,2,3,…r=1,2,3,\dots, is a square arrangement containing r2r^2 oranges.
    Find the number of oranges in layers 1313 to 2020 inclusive.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).