Sequences, recurrence relations and sigma notationEdexcel A-Level Maths: Revision notes

Section 1

Sequences and the nth term

A sequence is an ordered list of numbers $u_1,u_2,u_3,\ldots$ . It can be defined by a position-to-term (nth term) formula, such as $u_n=\frac{1}{3n+1}$ , which gives any term directly: $u_1=\frac14$ , $u_2=\frac17$ . Substituting $n=10$ finds $u_{10}$ without finding the earlier terms. A sequence is finite if it stops and infinite if it continues for ever.

To test whether a number is in the sequence, set $u_n$ equal to it and solve for $n$ : $n$ must be a positive integer.

Key termssequencenth termposition-to-term

Exam tip

$n$ must be a positive integer. If solving $u_n=k$ gives $n=4.5$ , then $k$ is not a term of the sequence.

Section 2

Recurrence relations

A recurrence relation (term-to-term rule) defines each term from the one before: $u_{n+1}=f(u_n)$ , together with a starting value such as $u_1=5$ . For $u_{n+1}=2u_n-3$ and $u_1=5$ : $u_2=7$ , $u_3=11$ , $u_4=19$ . You must work through the terms in order; you cannot jump to $u_{20}$ unless the sequence repeats or you can find a pattern.

A fixed point satisfies $u_{n+1}=u_n$ , so $x=f(x)$ . For $u_{n+1}=2u_n-3$ , $x=2x-3$ gives $x=3$ . If the sequence ever reaches $3$ it stays there for ever; starting from $5$ it moves away from $3$ .

Key termsrecurrence relationterm-to-term rulefixed point

Common mistake

Forgetting to state or use the first term. A recurrence relation without $u_1$ does not define a unique sequence.

Exam tip

Write each term with its working ( $u_3=2(7)-3=11$ ) so one slip does not carry through the later terms.

Section 3

Increasing, decreasing and periodic sequences

A sequence is increasing if $u_{n+1}>u_n$ for all $n$ , and decreasing if $u_{n+1}<u_n$ for all $n$ . To prove it, show the sign of $u_{n+1}-u_n$ . For $u_n=\frac{1}{3n+1}$ : $u_{n+1}=\frac{1}{3n+4}<\frac{1}{3n+1}=u_n$ , so it is decreasing. For $u_n=2^n$ : $u_{n+1}=2\times2^n>u_n$ , so it is increasing.

A sequence is periodic if the terms repeat in a cycle: $u_{n+k}=u_n$ for all $n$ . The smallest such $k$ is the order. For $u_{n+1}=\frac{1}{u_n}$ with $u_1=3$ : $u_2=\frac13$ , $u_3=3$ , so it is periodic of order $2$ .

A sequence can be none of these: $u_n=n^2-8n+20$ falls for $n\le3$ and then rises, because $u_{n+1}-u_n=2n-7$ changes sign.

Key termsincreasingdecreasingperiodicorder

Common mistake

Checking only the first few terms. To show a sequence is increasing or decreasing for all $n$ , work with $u_{n+1}-u_n$ .

Section 4

Sigma notation

$\sum_{r=1}^{n}u_r$ means $u_1+u_2+\cdots+u_n$ : the sum of the terms from $r=1$ up to $r=n$ . The lower limit is where $r$ starts and the upper limit is where it stops, so $\sum_{r=3}^{7}u_r$ has $7-3+1=5$ terms. For $u_r=r^2$ : $\sum_{r=1}^{4}r^2=1+4+9+16=30$ . A sum can also be taken over a recurrence sequence: generate the terms in order and add them.

Key termssigma notationlimits

Common mistake

Miscounting terms: $\sum_{r=m}^{n}$ has $n-m+1$ terms, not $n-m$ .

Section 5

Sums of constants and splitting sums

A constant added $n$ times gives $\sum_{r=1}^{n}1=n$ , so $\sum_{r=1}^{n}k=kn$ . Sums can be split and scaled: $\sum(a_r+b_r)=\sum a_r+\sum b_r$ and $\sum ka_r=k\sum a_r$ .

Worked example: $\sum_{r=1}^{7}(u_r-4)=\sum_{r=1}^{7}u_r-\sum_{r=1}^{7}4=\sum_{r=1}^{7}u_r-4\times7$ . With periodic sequences, group whole cycles: if $2,-1,\frac12$ repeats, each cycle sums to $\frac32$ , so $33$ cycles sum to $49.5$ .

Key termssum of constants

Exam tip

In a long sum of a periodic sequence, find the sum of one cycle, count complete cycles, then add any leftover terms.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Sequences, recurrence relations and sigma notation

A sequence is defined by $u_1=5$ and $u_{n+1}=2u_n-3$ for $n\ge1$ .
Calculate $\sum_{r=1}^{4}u_r$ .2 marks
A sequence is defined by $u_1=2$ and $u_{n+1}=\frac{8}{u_n}$ for $n\ge1$ .
Explain why the sequence is periodic and state its order.2 marks
A sequence has $n$ th term $u_n=n^2-8n+20$ for $n\ge1$ .
Show that $u_{n+1}-u_n=2n-7$ , and hence find the values of $n$ for which $u_{n+1}<u_n$ .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).