Edexcel A-level Further Maths revision notes

Recurrence relations

Section 6

Years 1 and 2

2 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section 6

Checked against Edexcel 9FM0 section 6. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

In the exam: Formulae booklet provided · calculator allowed in every paper

Open the printable pack

6.1

Year 1

Use of recurrence relations to model appropriate problems.

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

A recurrence relation defines each term of a sequence in terms of earlier terms, together with enough starting values to get the sequence going.
It is the natural model whenever a quantity changes in a fixed way from one period to the next.
A first-order relation $u_{n+1}=au_n+b$ describes a quantity that is multiplied by $a$ and then increased by $b$ each period, which covers compound interest with regular deposits, a loan repaid by fixed instalments, a population with constant proportional growth and constant immigration, and a drug whose concentration decays by a fixed proportion between doses.
A second-order relation $u_{n+2}=au_{n+1}+bu_n+c$ appears when the next term depends on the two previous ones, as in counting problems where an arrangement of length $n+2$ is built by adding either one item to an arrangement of length $n+1$ or two items to one of length $n$ .
Setting up the model means defining the variable precisely, including its units and what $n$ counts, writing the relation, and stating the initial condition or conditions; a relation without its initial values does not determine the sequence.

Worked example

A savings account pays $4\%$ interest at the end of each year, after which $£500$ is deposited. The account starts with $£2000$ . Set up a recurrence relation and find the balance after three years.

1.Let $u_n$ be the balance in pounds at the end of year $n$ , with $u_0=2000$ .
2.Interest multiplies the balance by $1.04$ , then $£500$ is added, so $u_{n+1}=1.04u_n+500$ .
3. $u_1=1.04\times2000+500=2580$ .
4. $u_2=1.04\times2580+500=3183.20$ .
5. $u_3=1.04\times3183.20+500=3810.53$ to the nearest penny.

Answer: $u_{n+1}=1.04u_n+500$ with $u_0=2000$ ; the balance after three years is $£3810.53$ .

Common mistakes

Don't fall into the trap of writing $u_{n+1}=0.04u_n+500$ , which adds only the interest and loses the capital.
Don't fall into the trap of giving the relation without stating an initial value, so the sequence is not determined.
Don't fall into the trap of leaving the variable undefined, so it is unclear whether $u_n$ is the balance at the start or the end of year $n$ .

Exam tip

Define $u_n$ in a full sentence with its units and say what $n$ counts; the modelling mark is for the definition as much as for the relation.

Tier 1 · Easy

ORIGINAL

A colony of insects grows by

30\%

each week, and

200

further insects are introduced at the end of each week. There are

500

insects at the start. Write down a recurrence relation for the number of insects and state the initial condition.

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

A loan of

£8000

is charged interest at

5\%

at the end of each year, immediately after which a repayment of

£1500

is made. Write down a recurrence relation for the amount owed, and calculate the amount owed after three years.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

A loan of

£8000

is charged

5\%

interest at the end of each year, followed immediately by a repayment of

£1500

. Show that the amount owed after

n

years is

u_n=30000-22000(1.05)^n

, and find the year in which the loan is cleared and the size of the final repayment.

(8)

(Total for Question 1 is 8 marks)

Your progress and exam materials

This section: Evidence from your answers: 0/2 secureYour confidence: 0 self-rated secureTracker status: 0/2 secure, 0 shaky, 2 unseen

Overall: Evidence from your answers: 0/158 secureYour confidence: 0 self-rated secureTracker status: 0/158 secure, 0 shaky, 158 unseen

Progress is saved on this device for guests and accounts right now; cross-device account sync is not live yet.

6.2

Years 1 and 2

Solution of first and second order linear homogeneous and non-homogeneous recurrence relations.

Notes

Evidence from your answers: none yet

Your confidence:

Explanation

A linear recurrence relation is solved by adding a complementary function to a particular solution, exactly as for a differential equation. For a first-order relation $u_{n+1}=au_n+b$ with $a\neq1$ , the complementary function is $Aa^n$ and a constant particular solution $x$ satisfies $x=ax+b$ , so $x=\frac{b}{1-a}$ ; the initial value then fixes $A$ .
For a second-order relation $u_{n+2}=au_{n+1}+bu_n$ , substitute $u_n=m^n$ to obtain the auxiliary equation $m^2=am+b$ . If it has distinct roots $\alpha$ and $\beta$ the complementary function is $A\alpha^n+B\beta^n$ ; if it has a repeated root $\alpha$ the complementary function is $(A+Bn)\alpha^n$ .
For a non-homogeneous relation $u_{n+2}=au_{n+1}+bu_n+c$ with a constant term, try a constant particular solution; if a constant already solves the homogeneous equation, which happens when $m=1$ is a root, try a term in $n$ instead, and if $m=1$ is a repeated root try a term in $n^2$ .
The two initial values then determine $A$ and $B$ .
Always check the closed form by generating the first two or three terms from it and comparing them with the recurrence.

Worked example

Solve $u_{n+2}=5u_{n+1}-6u_n$ with $u_1=1$ and $u_2=4$ .

1.The auxiliary equation is $m^2-5m+6=0$ , so $m=2$ or $m=3$ .
2.The general solution is $u_n=A2^n+B3^n$ .
3. $u_1=1$ gives $2A+3B=1$ ; $u_2=4$ gives $4A+9B=4$ .
4.Subtracting twice the first from the second gives $3B=2$ , so $B=\frac23$ and $A=-\frac12$ .
5.Hence $u_n=-\frac12\times2^n+\frac23\times3^n=2\times3^{n-1}-2^{n-1}$ .

Answer: $u_n=2\times3^{n-1}-2^{n-1}$ , giving $1, 4, 14, 46,\dots$ as required.

Common mistakes

Don't fall into the trap of using $A\alpha^n+B\beta^n$ when the auxiliary equation has a repeated root, which cannot fit two initial conditions.
Don't fall into the trap of trying a constant particular solution when $m=1$ is a root of the auxiliary equation.
Don't fall into the trap of substituting the initial conditions into the complementary function alone, forgetting the particular solution.

Exam tip

Generate the first three terms from your closed form and check them against the recurrence; it takes twenty seconds and catches almost every sign error.

Tier 1 · Easy

ORIGINAL

Solve the recurrence relation

u_{n+1}=3u_n+2

with

u_1=1

, giving

u_n

in terms of

n

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

Solve the recurrence relation

u_{n+1}=4u_n-9

with

u_1=5

, and hence find

u_5

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

Solve the recurrence relation

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

with

u_1=3

and

u_2=10

, explaining carefully why the usual form of the complementary function has to be modified.

(7)

(Total for Question 1 is 7 marks)

See everything due today — your streak and what to revise next, kept on this device.

Want help turning these notes into marks?

Bring a tricky specification point or a recent answer, and we can work through the method and exam wording together.

Book a free intro call