Sigma Notation

Read and write sums in sigma notation, spot the AP or GP inside, and evaluate them, including sums that do not start at 1 and sums to infinity.

In the exam for Edexcel Further Pure Maths 4PM1 · 11 real past-paper questions below

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Reading Σ

Writing $2 + 5 + 8 + 11 + \dots$ all the way to the 50th term takes a whole line. Mathematicians use a short code instead. The code is the Greek capital letter sigma, $Σ$ , which is their letter S, for "sum".

Formula

r = 1 \sum n f (r)

$f (r)$ — the rule that makes each term, such as $3 r - 1$ .
$r = 1$ — the counter $r$ and the value it starts at.
$n$ — the value the counter stops at.

Read it as: "put $r = 1$ into the rule, then $r = 2$ , then $r = 3$ , and so on up to $n$ . Add all the results." The counter can be any letter, such as $r$ , $k$ or $j$ .

Worked example

1
Write out $\sum_{r = 1}^{5} (3 r - 1)$ and find its value.
2
$r = 1 : 3 (1) - 1 = 2$ — Put the start value into the rule.
3
$r = 2, 3, 4, 5 : 5, 8, 11, 14$ — Keep going, one whole number at a time, up to 5.
4
$2 + 5 + 8 + 11 + 14 = 40$ — Add all five terms.

Rule

Number of terms = top − bottom + 1. For example, $\sum_{r = 4}^{12}$ has $12 - 4 + 1 = 9$ terms. The +1 is there because both ends count.

Key idea

The Swiss mathematician Leonhard Euler began using $Σ$ for sums in the 1750s. Mathematicians around the world still use his symbol.

Key idea

$Σ$ is just a short way to say: list the terms from the bottom value to the top value, then add them.

Spot the AP or GP

To add up a long sigma sum, first find out what kind of series is inside. Look at how the counter appears in the rule.

Counter times a number: $\sum (4 r + 3)$

Each term goes up by 4. It is an arithmetic series with $d = 4$ .

Counter in a power: $\sum 5 (2)^{k - 1}$

Each term is 2 times the one before. It is a geometric series with ratio 2.

Rule

To find the first term, always put the bottom value of the counter into the rule. Then find the second term the same way to check $d$ or the ratio.

Worked example

1
Find the first term and common difference of the arithmetic series $\sum_{r = 1}^{n} (4 r + 3)$ .
2
$r = 1 : 4 + 3 = 7$ — First term a = 7.
3
$r = 2 : 8 + 3 = 11$ — Second term.
4
$d = 11 - 7 = 4$ — The common difference is the number in front of r.

Worked example

1
Find the first term and common ratio of the geometric series $\sum_{k = 1}^{n} 3 (2)^{k}$ .
2
$k = 1 : 3 \times 2^{1} = 6$ — First term a = 6, not 3.
3
$k = 2 : 3 \times 2^{2} = 12$ — Second term.
4
$ratio = 12 \div 6 = 2$ — The common ratio is the number being raised to the power.

Common mistake

Taking the number on its own as the first term. In $\sum_{r = 1}^{n} (4 r + 3)$ the first term is 7, not 3. In $\sum_{k = 1}^{n} 3 (2)^{k}$ it is 6, not 3.

Key idea

Put in the bottom value for the first term. Then the number in front of $r$ is $d$ , or the base of the power is the ratio.

Evaluate a sum from 1

Once you know the type of series, use its sum formula. Both are on the 4PM1 formulae sheet.

Formula

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

$a$ — the first term, from the bottom value.
$d$ — the common difference.
$n$ — the number of terms.

Formula

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

$a$ — the first term of the geometric series.
$r$ — here r is the common ratio, not the counter. Use a different counter letter, like k, to avoid a mix-up.

Worked example

1
Show that $\sum_{r = 1}^{n} (5 r - 2) = \frac{n}{2} (5 n + 1)$ .
2
$a = 5 (1) - 2 = 3, d = 5$ — First term from r = 1. The number in front of r is d.
3
$S_{n} = \frac{n}{2} [6 + 5 (n - 1)]$ — Put a and d into the sum formula. Show this line: it earns the method mark.
4
$= \frac{n}{2} (5 n + 1)$ — Simplify inside the bracket: 6 + 5n − 5 = 5n + 1.

Worked example

1
Given that $\sum_{r = 1}^{n} (5 r - 2) = 366$ , find $n$ .
2
$\frac{n}{2} (5 n + 1) = 366$ — Use the result you just showed.
3
$5 n^{2} + n - 732 = 0$ — Multiply by 2 and expand.
4
$(n - 12) (5 n + 61) = 0$ — Factorise, or use the quadratic formula.
5
$n = 12$ — n must be a positive whole number, so reject the negative root.

Worked example

1
Evaluate $\sum_{k = 1}^{6} 2 (3)^{k}$ .
2
$a = 2 \times 3 = 6, ratio = 3, n = 6$ — First term from k = 1. The base of the power is the ratio.
3
$S_{6} = \frac{6 ( 3 ^{6} - 1 )}{3 - 1}$ — The ratio is bigger than 1, so multiply the top and bottom of the formula by −1. You get the same sum, with no minus signs.
4
$= 3 \times 728 = 2184$ — 3⁶ = 729, and 6 ÷ 2 = 3.

Common mistake

Answering a "show that" by working out a few terms. You must use the sum formula with $a$ and $d$ , and reach the exact expression given.

Key idea

Find the first term, then $d$ or the ratio, then the number of terms. After that it is just the sum formula.

Sums that start later

Not every sum starts at 1. To add terms 21 to 40, take the sum up to 40, then remove the terms you do not want: the first 20.

Formula

r = k \sum m f (r) = S_{m} - S_{k - 1}

$m$ — the top value. Add everything up to here.
$k$ — the bottom value. Remove everything before it, which is up to $k - 1$ .

Worked example

1
Evaluate $\sum_{r = 21}^{40} (5 r - 2)$ .
2
$S_{40} - S_{20}$ — Terms 21 to 40: all 40 terms, minus the first 20.
3
$S_{40} = \frac{40}{2} (5 \times 40 + 1) = 4020$ — Use the result you showed in the last step, $S_{n} = \frac{n}{2} (5 n + 1)$ , with n = 40.
4
$S_{20} = \frac{20}{2} (5 \times 20 + 1) = 1010$ — Same formula with n = 20.
5
$4020 - 1010 = 3010$ — Subtract.

You can also add the block directly. Find its first term, its last term, and how many terms it has.

Worked example

1
Check the same sum directly.
2
$r = 21 : 103, r = 40 : 198$ — First and last terms of the block.
3
$40 - 21 + 1 = 20 terms$ — Top − bottom + 1.
4
$\frac{20}{2} (103 + 198) = 3010$ — The same answer.

Worked example

1
Evaluate $\sum_{k = 4}^{9} 2^{k}$ .
2
$a = 2^{4} = 16, ratio = 2$ — First term from the bottom value, k = 4.
3
$9 - 4 + 1 = 6 terms$ — Count the terms.
4
$S = \frac{16 ( 2 ^{6} - 1 )}{2 - 1} = 16 \times 63 = 1008$ — GP sum with n = 6.

Common mistake

Writing $S_{40} - S_{21}$ . That also removes the 21st term, which you need. Subtract the sum up to one before the bottom value.

Key idea

Sum from $k$ to $m$ : everything up to $m$ , minus everything up to $k - 1$ .

Write a sum in Σ form

Sometimes you go the other way: you have a series and must write it with $Σ$ . You need two things: a rule for the rth term, and the number of terms.

Worked example

1
Write $8 + 13 + 18 + \dots + 98$ in sigma notation.
2
$a = 8, d = 5$ — Each term is 5 more than the one before.
3
$u_{r} = 8 + 5 (r - 1) = 5 r + 3$ — nth term formula, simplified. Check: r = 1 gives 8.
4
$5 r + 3 = 98 \Rightarrow r = 19$ — Find which term is the last one.
5
$\sum_{r = 1}^{19} (5 r + 3)$ — Rule inside, from 1 up to 19.

Worked example

1
Write the first 10 terms of the geometric series $6 + 9 + 13.5 + \dots$ in sigma notation.
2
$a = 6, ratio = \frac{3}{2}$ — 9 ÷ 6 = 1.5.
3
$u_{k} = 6 (\frac{3}{2})^{k - 1}$ — nth term of a geometric series: first term times ratio to the power k − 1.
4
$\sum_{k = 1}^{10} 6 (\frac{3}{2})^{k - 1}$ — Rule inside, from 1 up to 10.
5
$= \sum_{k = 1}^{10} 4 (\frac{3}{2})^{k}$ — The same sum: 6 ÷ 1.5 = 4, so one extra power of 1.5 is balanced by 4 in front.

An $\infty$ on top means the sum to infinity. It exists only when the ratio is between −1 and 1.

Worked example

1
Evaluate $\sum_{k = 1}^{\infty} 12 (\frac{1}{3})^{k}$ .
2
$a = 12 \times \frac{1}{3} = 4, ratio = \frac{1}{3}$ — First term from k = 1.
3
$\frac{1}{3} < 1$ — The ratio is between −1 and 1, so the sum to infinity exists.
4
$S_{\infty} = \frac{4}{1 - \frac{1}{3}} = 6$ — Use a ÷ (1 − ratio).

Common mistake

Using the first term as the number in the rule. For $8 + 13 + 18 + \dots$ the rule is $5 r + 3$ , not $5 r + 8$ . Always check by putting $r = 1$ in.

Key idea

To write a Σ: find the rule for the rth term, check it with $r = 1$ , then find which term is the last.

Which method? Read the question

The shape of the sum and the wording tell you what to do.

You see	You do
Counter times a number	Arithmetic, like $\sum (4 r + 3)$ : first term from the bottom value, $d$ = number in front
Counter in a power	Geometric, like $\sum 3 (2)^{k}$ : first term from the bottom value, ratio = base
"Show that the sum = …"	Find $a$ and $d$ (or the ratio), use the sum formula, simplify
Sum equals a number, find $n$	Quadratic in $n$ , keep the positive whole answer
Bottom value is not 1	$S_{m} - S_{k - 1}$ , or add the block directly
$\infty$ on top	Check the ratio is between −1 and 1, then $\frac{a}{1 - r}$
"Write in sigma notation"	Rule for the rth term, then the number of terms

Common mistake

Off-by-one slips: the number of terms is top − bottom + 1, and a sum from $k$ subtracts $S_{k - 1}$ .

Key idea

Write out the first two terms every time. They tell you the type, the first term and the step.

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