Summation of Series

This lesson covers the summation of series using standard results, sigma notation, and algebraic techniques. Summation is a key topic in AQA Further Mathematics and requires fluency with standard formulae and the ability to decompose more complex sums.

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Sigma Notation

The symbol $\sum$∑ (capital sigma) denotes summation:

$\sum_{r=1}^{n} f(r) = f(1) + f(2) + \cdots + f(n)$∑r=1n​f(r)=f(1)+f(2)+⋯+f(n)

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Standard Results

You must memorise and apply these formulae:

Sum	Formula
$\sum_{r=1}^{n} 1$∑r=1n​1	$n$n
$\sum_{r=1}^{n} r$∑r=1n​r	$\frac{n(n+1)}{2}$2n(n+1)​
$\sum_{r=1}^{n} r^2$∑r=1n​r2	$\frac{n(n+1)(2n+1)}{6}$6n(n+1)(2n+1)​
$\sum_{r=1}^{n} r^3$∑r=1n​r3	$\left(\frac{n(n+1)}{2}\right)^2$(2n(n+1)​)2

Key Identity: $\sum r^3 = \left(\sum r\right)^2$∑r3=(∑r)2. This elegant result is sometimes useful in proofs.

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Properties of Sums

Property	Formula
Constant factor	$\sum_{r=1}^{n} cf(r) = c\sum_{r=1}^{n} f(r)$∑r=1n​cf(r)=c∑r=1n​f(r)
Sum/difference	$\sum_{r=1}^{n} (f(r) \pm g(r)) = \sum f(r) \pm \sum g(r)$∑r=1n​(f(r)±g(r))=∑f(r)±∑g(r)
Shifting limits	$\sum_{r=k}^{n} f(r) = \sum_{r=1}^{n} f(r) - \sum_{r=1}^{k-1} f(r)$∑r=kn​f(r)=∑r=1n​f(r)−∑r=1k−1​f(r)

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Worked Examples

Example 1: Find $\sum_{r=1}^{20} (3r^2 + 2r - 1)$∑r=120​(3r2+2r−1).

$= 3\sum_{r=1}^{20} r^2 + 2\sum_{r=1}^{20} r - \sum_{r=1}^{20} 1$=3∑r=120​r2+2∑r=120​r−∑r=120​1

$= 3 \cdot \frac{20 \cdot 21 \cdot 41}{6} + 2 \cdot \frac{20 \cdot 21}{2} - 20$=3⋅620⋅21⋅41​+2⋅220⋅21​−20

$= 3 \cdot 2870 + 2 \cdot 210 - 20 = 8610 + 420 - 20 = 9010$=3⋅2870+2⋅210−20=8610+420−20=9010

Example 2: Find $\sum_{r=1}^{n} r(r+1)$∑r=1n​r(r+1).

$\sum_{r=1}^{n} r(r+1) = \sum_{r=1}^{n} (r^2 + r) = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$∑r=1n​r(r+1)=∑r=1n​(r2+r)=6n(n+1)(2n+1)​+2n(n+1)​

$= \frac{n(n+1)(2n+1) + 3n(n+1)}{6} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}$=6n(n+1)(2n+1)+3n(n+1)​=6n(n+1)(2n+4)​=3n(n+1)(n+2)​

Example 3: Find $\sum_{r=10}^{30} r$∑r=1030​r.

$= \sum_{r=1}^{30} r - \sum_{r=1}^{9} r = \frac{30 \cdot 31}{2} - \frac{9 \cdot 10}{2} = 465 - 45 = 420$=∑r=130​r−∑r=19​r=230⋅31​−29⋅10​=465−45=420

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Summing Polynomial Expressions

For any polynomial expression in $r$r, expand and use the standard results:

$\sum_{r=1}^{n} (ar^3 + br^2 + cr + d) = a\sum r^3 + b\sum r^2 + c\sum r + dn$∑r=1n​(ar3+br2+cr+d)=a∑r3+b∑r2+c∑r+dn

Always simplify and factorise your final answer.

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Practice Problems

1. Find $\sum_{r=1}^{n} (2r - 1)$∑r=1n​(2r−1) and simplify.
2. Find $\sum_{r=1}^{50} r^2$∑r=150​r2.
3. Evaluate $\sum_{r=1}^{n} r(r + 2)$∑r=1n​r(r+2).
4. Find $\sum_{r=5}^{20} r^2$∑r=520​r2.

Solutions: