Edexcel A-level Further Maths revision notes

Further sequences and series (Further Pure 2)

Section 6

Years 1 and 2

3 specification points

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

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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.

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6.1

Years 1 and 2

First and second order recurrence relations.

Notes

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Explanation

A recurrence relation defines each term of a sequence from earlier terms together with enough initial values to start it. A first order relation uses only the previous term, as in $u_{n+1}=au_{n}+b$ , and needs one initial value.
A second order relation uses the two previous terms, as in $u_{n+2}=pu_{n+1}+qu_{n}$ , and needs two.
A relation is homogeneous when every term involves the sequence, so $u_{n+2}-5u_{n+1}+6u_{n}=0$ is homogeneous while $u_{n+2}-5u_{n+1}+6u_{n}=4$ is not.
Modelling questions ask you to build the relation: identify what one step does to the current value, translate multiplication into the coefficient and any fixed addition into the constant term.
For a homogeneous linear relation the auxiliary equation is obtained by trying $u_{n}=\lambda^{n}$ , which gives $\lambda^{2}=p\lambda+q$ after dividing by $\lambda^{n}$ ; its roots control the behaviour of the sequence and are the first step towards a closed form.

Worked example

A sequence is defined by $u_{n+2}=2u_{n+1}+3u_{n}$ with $u_{0}=1$ and $u_{1}=3$ . Find $u_{2}$ , $u_{3}$ and $u_{4}$ , and write down the auxiliary equation.

1. $u_{2}=2\times3+3\times1=9$ .
2. $u_{3}=2\times9+3\times3=27$ .
3. $u_{4}=2\times27+3\times9=81$ .
4.Trying $u_{n}=\lambda^{n}$ gives $\lambda^{2}=2\lambda+3$ , that is $\lambda^{2}-2\lambda-3=0$ .

Answer: $u_{2}=9$ , $u_{3}=27$ , $u_{4}=81$ , and the auxiliary equation is $\lambda^{2}-2\lambda-3=0$ .

Common mistakes

Don't fall into the trap of supplying only one initial value for a second order relation, which leaves the sequence undetermined.
Don't fall into the trap of reading $u_{n+2}=pu_{n+1}+qu_{n}$ off with the coefficients swapped when forming the auxiliary equation.
Don't fall into the trap of adding the fixed amount before applying the growth factor in a modelling question whose wording says otherwise.

Exam tip

Write the first few terms out explicitly even when the question only asks for the relation; they are the quickest check that your coefficients are the right way round.

Tier 1 · Easy

ORIGINAL

A sequence is defined by

u_{n+1}=3u_{n}-2

with

u_{1}=4

. Find

u_{2}

u_{3}

and

u_{4}

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

An account starts with

\pounds1000

. At the end of each year interest of

4\%

is added and then a further

\pounds200

is deposited. Write down a recurrence relation for the amount

A_{n}

pounds in the account after

n

years, and find

A_{3}

to the nearest penny.

(6)

(Total for Question 1 is 6 marks)

Tier 3 · Hard

ORIGINAL

Three pegs stand in a row and

n

discs of different sizes are stacked on the first peg, largest at the bottom. A move transfers the top disc of one peg to another peg, and a disc may never be placed on a smaller disc. Let

h_{n}

be the least number of moves needed to transfer the whole stack to another peg. Explain why

h_{n}=2h_{n-1}+1

for

n\geqslant2

, and find

h_{6}

given that

h_{1}=1

(7)

(Total for Question 1 is 7 marks)

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6.2

Year 1

The solution of recurrence relations to obtain closed forms.

Notes

Evidence from your answers: none yet

Your confidence:

Explanation

For a first order relation $u_{n+1}=au_{n}+b$ with $a\ne1$ , find the fixed point $c$ from $c=ac+b$ , so $c=\dfrac{b}{1-a}$ ; then $u_{n}-c$ is geometric with ratio $a$ and $u_{n}=c+a^{n}\left(u_{0}-c\right)$ .
For a homogeneous second order relation $u_{n+2}=pu_{n+1}+qu_{n}$ , solve the auxiliary equation $\lambda^{2}-p\lambda-q=0$ .
Distinct real roots $\alpha$ and $\beta$ give $u_{n}=A\alpha^{n}+B\beta^{n}$ ; a repeated root $\alpha$ gives $u_{n}=\left(A+Bn\right)\alpha^{n}$ .
The constants $A$ and $B$ are then fixed by substituting the two initial values and solving simultaneously.
For a non-homogeneous relation, the general solution is the complementary function, found as above, plus any particular solution: try a constant when the right-hand side is constant, and try $k\mu^{n}$ when it is $\mu^{n}$ , unless $\mu$ is already a root of the auxiliary equation, in which case multiply the trial by $n$ .

Worked example

Solve $u_{n+2}=5u_{n+1}-6u_{n}$ with $u_{0}=1$ and $u_{1}=4$ .

1.The auxiliary equation is $\lambda^{2}-5\lambda+6=0$ , with roots $\lambda=2$ and $\lambda=3$ .
2.So $u_{n}=A2^{n}+B3^{n}$ .
3. $u_{0}=1$ gives $A+B=1$ ; $u_{1}=4$ gives $2A+3B=4$ .
4.Subtracting twice the first from the second gives $B=2$ , hence $A=-1$ .

Answer: $u_{n}=2\times3^{n}-2^{n}$ .

Common mistakes

Don't fall into the trap of writing $u_{n}=A\alpha^{n}+B\alpha^{n}$ for a repeated root, which collapses to a single constant.
Don't fall into the trap of fitting the constants using $u_{1}$ and $u_{2}$ when the given initial values are $u_{0}$ and $u_{1}$ .
Don't fall into the trap of forgetting the particular solution when the relation has a non-zero right-hand side.

Exam tip

Always test your closed form on the next term the recurrence produces; one substitution catches almost every sign or constant error.

Tier 1 · Easy

ORIGINAL

Find a closed form for the sequence defined by

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

with

u_{0}=5

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

Find a closed form for the sequence defined by

u_{n+1}=3u_{n}-2

with

u_{1}=4

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

Find a closed form for the sequence defined by

u_{n+1}=2u_{n}+3

with

u_{0}=1

(6)

(Total for Question 1 is 6 marks)

6.3

Year 1

Proof by induction of closed forms.

Notes

Evidence from your answers: none yet

Your confidence:

Explanation

A proof by induction of a closed form has four written parts. State the proposition $P(n)$ precisely.
Prove the base case by evaluating both the recurrence's initial value and the closed form at the smallest index. Assume $P(k)$ for some $k$ in range, then deduce $P(k+1)$ by substituting the assumed closed form into the recurrence relation and simplifying to the closed form with $k+1$ in place of $k$ .
Finish with the conclusion, naming both the base case and the inductive step.
A second order recurrence needs two base cases and the assumption of both $P(k)$ and $P(k+1)$ , because the step uses two previous terms; this is two-step ordinary induction, not strong induction, which would assume $P(m)$ for every $m$ up to $k$ .
The algebra in the inductive step is nearly always a matter of factorising out the common power, so aim for that form early and keep the index arithmetic explicit.

Worked example

A sequence is defined by $u_{1}=4$ and $u_{n+1}=3u_{n}-2$ . Prove by induction that $u_{n}=3^{n}+1$ for all integers $n\ge1$ .

1.Base case: $3^{1}+1=4=u_{1}$ , so the statement holds for $n=1$ .
2.Assume $u_{k}=3^{k}+1$ for some integer $k\ge1$ .
3.Then $u_{k+1}=3u_{k}-2=3\left(3^{k}+1\right)-2=3^{k+1}+3-2$ .
4. $=3^{k+1}+1$ , which is the statement for $n=k+1$ .

Answer: True for $n=1$ and inherited from each $k$ to $k+1$ , so $u_{n}=3^{n}+1$ for all integers $n\ge1$ by induction.

Common mistakes

Don't fall into the trap of assuming the result for all $n$ rather than for one particular $k$ , which assumes what is to be proved.
Don't fall into the trap of giving only one base case for a second order recurrence, where two are needed.
Don't fall into the trap of omitting the closing statement that names the base case and the inductive step.

Exam tip

Write the target expression for $n=k+1$ down before you start simplifying; knowing where you are heading makes the factorisation obvious.

Tier 1 · Easy

ORIGINAL

A sequence is defined by

u_{1}=2

and

u_{n+1}=4u_{n}-3

. Prove by induction that

u_{n}=4^{n-1}+1

for all integers

n\geqslant1

(5)

(Total for Question 1 is 5 marks)

Tier 2 · Standard

ORIGINAL

A sequence is defined by

u_{1}=1

and

u_{n+1}=3u_{n}+2

. Prove by induction that

u_{n}=2\times3^{n-1}-1

for all integers

n\ge1

(6)

(Total for Question 1 is 6 marks)

Tier 3 · Hard

ORIGINAL

A sequence is defined by

u_{0}=1

u_{1}=4

and

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

. Prove by induction that

u_{n}=2\times3^{n}-2^{n}

for all integers

n\ge0

(8)

(Total for Question 1 is 8 marks)

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