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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn9FM0Option · Decision Mathematics 2Recurrence relationsSolving recurrence relations

Recurrence relations · Subtopic 1 of 1

Recurrence Relations

A recurrence defines each term from earlier ones. First order: $u_{n+1}=ku_n+c$ has solution $Ak^n+$ constant. Second order: $u_{n+2}+pu_{n+1}+qu_n=0$ is solved with the auxiliary equation $r^2+pr+q=0$ , exactly like differential equations but with powers instead of exponentials.

Edexcel 9FM0 Option≈ 45 minRevision 1 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Form recurrence relations to model situations.
Solve first-order linear recurrences.
Solve second-order recurrences using the auxiliary equation.
Use initial conditions to find constants.

Pause and think

£1000 earns 5% interest a year and you add £100 each year. How do you describe the balance?

Section 2 of 14

Core idea

Start hereRabbits and repayments

Fibonacci's rabbits, a growing savings account and a shrinking loan have one thing in common: each value is built from the previous ones. Working out term 100 step by step is painful. A closed formula gets you there in one line.

The big idea

Linear recurrences with constant coefficients are solved like differential equations, with $r^n$ in place of $e^{mx}$ . The auxiliary equation gives the general solution, a particular solution handles constants, and initial terms fix the constants.

1. First-order recurrences

Solving

u_{n+1}=ku_n+c

$\lambda=k\lambda+c\Rightarrow\lambda=\frac c{1-k}$
Look for a constant solution ( $k\ne1$ ).
$v_n=u_n-\lambda$
Measure from that constant.
$v_{n+1}=kv_n$
Subtract the two equations; the $c$ cancels.
$u_n=Ak^n+\frac c{1-k}$
Geometric sequence plus the constant.

Worked example $u_{n+1}=2u_n+3$ , $u_0=1$

Find $u_n$ .

Goal: Closed form.
Strategy: Constant solution, then fit $A$ .

$\lambda=2\lambda+3\Rightarrow\lambda=-3$
Particular solution.
$u_n=A\cdot2^n-3$
General solution.
$1=A-3\Rightarrow A=4$
Use $u_0=1$ .

Answer

u_n=4\cdot2^n-3

✓ Check

u_1=5

from both the recurrence and the formula ✓.

2. Second-order recurrences

Trying

u_n=r^n

$r^{n+2}+pr^{n+1}+qr^n=0$
Substitute into $u_{n+2}+pu_{n+1}+qu_n=0$ .
$r^2+pr+q=0$
Divide by $r^n\ne0$ .
$u_n=A\alpha^n+B\beta^n$
Two roots give two independent solutions; linearity lets us add them.

Repeated root

\alpha

u_n=(A+Bn)\alpha^n

Worked exampleDistinct roots

Solve $u_{n+2}=5u_{n+1}-6u_n$ , $u_0=1$ , $u_1=4$ .

Goal: Closed form.
Strategy: Auxiliary equation, then two conditions.

$r^2-5r+6=0\Rightarrow r=2,3$
Factorise.
$u_n=A2^n+B3^n$
General solution.
$A+B=1,\ 2A+3B=4$
$n=0$ and $n=1$ .
$B=2,\ A=-1$
Solve simultaneously.

Answer

u_n=2\cdot3^n-2^n

✓ Check

u_2=5(4)-6(1)=14

and

2(9)-4=14

✓.

Method

Drawing diagram…

The same steps as second-order differential equations.

Common misconceptionFit constants last Apply initial conditions to the full solution (including any particular solution), never to the complementary part alone.

Edge caseWhen $k=1$

u_{n+1}=u_n+c

is arithmetic:

u_n=u_0+cn

. The constant-solution trick fails because

\lambda=\lambda+c

has no solution.

Pause and think

Why do Fibonacci ratios tend to $1.618\ldots$ ?

Summary

First order: $Ak^n+\frac c{1-k}$ .
Second order: $r^2+pr+q=0$ .
Repeated root: $(A+Bn)\alpha^n$ .
Check with a computed term.

Recurrence relation: Rule giving a term from earlier terms.
Auxiliary equation: Polynomial in $r$ from trying $u_n=r^n$ .
Closed form: Formula for $u_n$ directly in terms of $n$ .

HistoryFibonacci. Leonardo of Pisa (Fibonacci) posed the rabbit problem in 1202, giving

F_{n+2}=F_{n+1}+F_n

. Binet found its closed form in 1843.

Fun factGolden ratio. Ratios of consecutive Fibonacci numbers approach

\frac{1+\sqrt5}2\approx1.618

, the larger root of

r^2=r+1

Section 3 of 14

Key formulae

First order

u_{n+1}=ku_n+c\ (k\ne1):\ u_n=Ak^n+\frac c{1-k}

Second order

u_{n+2}+pu_{n+1}+qu_n=0:\ r^2+pr+q=0

Solutions

u_n=A\alpha^n+B\beta^n\ (\alpha\ne\beta);\quad u_n=(A+Bn)\alpha^n\ (\text{repeated})

TrickSame as DEs. Swap

e^{mx}

for

r^n

: auxiliary equations, repeated roots and particular solutions all work the same way.

Section 4 of 14

Worked examples

Example 1 · First order0 / 3 steps

Solve

u_{n+1}=2u_n+3

u_0=1

Example 2 · Second order0 / 3 steps

Solve

u_{n+2}=5u_{n+1}-6u_n

u_0=1

u_1=4

Section 5 of 14

Explore the graph

Terms approaching the golden ratio

Loading the interactive graph…

Try this

Fibonacci numbers grow like $\frac{\varphi^n}{\sqrt5}$ .
Each step multiplies by roughly $1.618$ .

The dominant root controls long-term growth.

In the real worldLoans. A loan with monthly interest and fixed repayments satisfies

u_{n+1}=(1+r)u_n-P

; solving it gives the repayment needed to clear the debt in a set time.

Section 6 of 14

Common mistakes

IncorrectUsing

u_1

u_0

CorrectCheck the starting index

A

changes if the first term is

u_1

IncorrectApplying conditions to the homogeneous part only

CorrectUse the full solution

The particular solution contributes to every term.

IncorrectRepeated root

A\alpha^n+B\alpha^n

Correct

(A+Bn)\alpha^n

You need two independent solutions.

Calculate the 100th term by hand

Closed form: hold my auxiliary equation.

Section 7 of 14

Exam tip

Exam tipVerify with a term. After solving, compute

u_2

from the recurrence and from your formula. They must match.

Section 8 of 14

Guided practice

SA-FP-RR-G011 markL1 FoundationNon-calculatorFirst order

u_{n+1}=2u_n+3

u_1=1

. Find

u_3

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-FP-RR-0012 marksL1 FoundationNon-calculatorFirst order

u_{n+1}=3u_n

u_0=2

. Find

u_4

Salam Academy Original Practice · Proposed mark scheme

SA-FP-RR-0022 marksL2 StandardNon-calculatorFirst order

Find the constant solution

u_n=k

u_{n+1}=2u_n+3

Salam Academy Original Practice · Proposed mark scheme

SA-FP-RR-0032 marksL2 StandardNon-calculatorSecond order

Find the roots of the auxiliary equation for

u_{n+2}=5u_{n+1}-6u_n

Salam Academy Original Practice · Proposed mark scheme

SA-FP-RR-0044 marksL3 ExamNon-calculatorSecond order

u_{n+2}=5u_{n+1}-6u_n

with

u_0=1

u_1=4

. Find

u_3

using the closed form.

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-FP-RR-0052 marksL3 ExamCalculatorSecond order

Find the larger root of the auxiliary equation for

u_{n+2}=u_{n+1}+u_n

(to 4 s.f.).

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

First order: $Ak^n+$ constant.
Second order: auxiliary equation.
Repeated root: $(A+Bn)\alpha^n$ .
Fit constants to initial terms.

Section 12 of 14

Quick quiz

Q1 · SA-FP-RR-Q011 markL1 FoundationNon-calculatorSecond order

u_{n+2}-4u_{n+1}+4u_n=0

has general solution

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: A level Further Mathematics 9FM0
Where: Decision 2 · Recurrence relations (also FP2 sequences)
Expected: Form and solve first- and second-order linear recurrence relations with constant coefficients, using auxiliary equations and particular solutions; prove closed forms by induction.
Calculator: Permitted; show enough working to earn the method marks.
Typical style: Edexcel often opens with a routine part and builds to a "show that" or "prove" part.
Open in: 9FM0 · Option

Section 14 of 14

Past paper questions

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Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel 9FM0, First teaching 2017 (official specification).