What is arithmetic slips in iteration?

Each new term depends on the previous, so one error propagates; work carefully.

SEAB Original-style practice: A sequence is defined by u_1 = 2 and u_n+1 = 12u_n + 3. Find the first four terms and the limit of the sequence as n → ∞.

u_1 = 2. u_2 = 12(2) + 3 = 4. u_3 = 12(4) + 3 = 5. u_4 = 12(5) + 3 = 5.5. The terms increase toward a limit. If u_n → L, then u_n+1 → L too, so L = 12L + 3, giving 12L = 3, so L = 6. Markers reward correctly iterating the first four terms, setting u_n+1 = u_n = L in the limit, and solving for L = 6.

SingaporeMathsSyllabus dot point

How do recurrence relations define a sequence, and how do we find or verify a closed form?

Use recurrence relations to generate sequences, find and verify a conjectured formula for the nth term, and analyse long-term behaviour

A focused answer to the H2 Mathematics outcome on recurrence relations. Generating terms from a recurrence, conjecturing and verifying a closed form, finding a limiting value, and recognising arithmetic and geometric recurrences.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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Examples in context
Try this

What this dot point is asking

SEAB wants you to use a recurrence relation (a rule giving each term from previous ones) to generate a sequence, conjecture and verify a closed-form expression for the $n$ th term, find a limiting value where one exists, and recognise when a recurrence produces an arithmetic or geometric progression.

The answer

What a recurrence relation is

A recurrence relation defines a sequence by giving one or more starting terms and a rule for the next term in terms of earlier ones, for example $u_{n+1} = \mathrm{f}(u_n)$ with $u_1$ specified. Generating terms is just repeated substitution (iteration).

Recognising standard types

If $u_{n+1} = u_n + d$ , the sequence is arithmetic with common difference $d$ .
If $u_{n+1} = r u_n$ , it is geometric with common ratio $r$ .

Mixed linear recurrences $u_{n+1} = au_n + b$ ( $a \neq 1$ ) combine a geometric part with a constant and tend to a limit when $|a| < 1$ .

Finding a limiting value

If the sequence converges to a limit $L$ , then both $u_n$ and $u_{n+1}$ approach $L$ . Substitute $L$ for both in the recurrence and solve the resulting equation. For $u_{n+1} = au_n + b$ this gives $L = \dfrac{b}{1 - a}$ , valid when $|a| < 1$ so the sequence actually converges.

Verifying a conjectured formula

To confirm a proposed $u_n$ :

Check it gives the correct first term(s).
Substitute it into the recurrence and show it reproduces $u_{n+1}$ .

A full proof for all $n$ uses mathematical induction, but verification of the initial condition plus the recurrence step is the routine algebra checked here.

Worked example

A sequence is defined by $u_1 = 1$ and $u_{n+1} = \dfrac{4u_n + 2}{u_n + 3}$ . Show the sequence increases toward a limit and find that limit.

Step 1: Generate a few terms

$u_1 = 1$ . $u_2 = \dfrac{4(1) + 2}{1 + 3} = \dfrac{6}{4} = 1.5$ . $u_3 = \dfrac{4(1.5) + 2}{1.5 + 3} = \dfrac{8}{4.5} \approx 1.78$ . The terms are increasing.

Step 2: Assume a limit and substitute

If $u_n \to L$ then $u_{n+1} \to L$ , so

L = \frac{4L + 2}{L + 3}

Step 3: Form a quadratic

Multiply through: $L(L + 3) = 4L + 2$ , so $L^2 + 3L = 4L + 2$ , giving $L^2 - L - 2 = 0$ .

Step 4: Solve and select the valid root

$(L - 2)(L + 1) = 0$ , so $L = 2$ or $L = -1$ . The terms are positive and increasing from $1$ , so the relevant limit is $L = 2$ .

Step 5: State the result

The sequence increases and converges to the limit $2$ .

Examples in context

Example 1. A discrete population model. A fish stock with $u_{n+1} = 1.1 u_n - 200$ (10% growth minus a fixed catch) has limiting level $L = \dfrac{-200}{1 - 1.1} = 2000$ only if $|a| < 1$ ; here $a = 1.1 > 1$ , so it diverges, warning that the catch is unsustainable below $2000$ .

Example 2. Iterating toward a root. The recurrence $x_{n+1} = \dfrac{1}{2}\left(x_n + \dfrac{2}{x_n}\right)$ converges to $\sqrt{2}$ : setting $x_{n+1} = x_n = L$ gives $L = \dfrac{1}{2}(L + \frac{2}{L})$ , so $L^2 = 2$ . This is the classic square-root iteration.

Try this

Q1. Generate the first four terms of $u_1 = 5$ , $u_{n+1} = 2u_n - 3$ . [2 marks]

Cue. $5, 7, 11, 19$ .

Q2. A sequence has $u_{n+1} = 0.4u_n + 6$ . Find its limit. [3 marks]

Cue. $L = 0.4L + 6$ , so $0.6L = 6$ , $L = 10$ .

Q3. State whether $u_{n+1} = u_n + 7$ defines an arithmetic or geometric progression, and give the relevant parameter. [1 mark]

Cue. Arithmetic, common difference $7$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA sequence is defined by

u_1 = 2

and

u_{n+1} = \dfrac{1}{2}u_n + 3

. Find the first four terms and the limit of the sequence as

n \to \infty

Show worked answer →

$u_1 = 2$ . $u_2 = \tfrac{1}{2}(2) + 3 = 4$ . $u_3 = \tfrac{1}{2}(4) + 3 = 5$ . $u_4 = \tfrac{1}{2}(5) + 3 = 5.5$ .

The terms increase toward a limit. If $u_n \to L$ , then $u_{n+1} \to L$ too, so $L = \tfrac{1}{2}L + 3$ , giving $\tfrac{1}{2}L = 3$ , so $L = 6$ .

Markers reward correctly iterating the first four terms, setting $u_{n+1} = u_n = L$ in the limit, and solving for $L = 6$ .

Original4 marksA sequence satisfies

u_{n+1} = 3u_n - 4

with

u_1 = 3

. Show that

u_n = 2 + 3^{n-1}

satisfies both the initial condition and the recurrence.

Show worked answer →

Initial condition: $u_1 = 2 + 3^0 = 2 + 1 = 3$ . Correct.

Recurrence check: with $u_n = 2 + 3^{n-1}$ ,
$3u_n - 4 = 3(2 + 3^{n-1}) - 4 = 6 + 3^n - 4 = 2 + 3^n = 2 + 3^{(n+1)-1} = u_{n+1}$ .

Both conditions hold, so the proposed formula is correct.

Markers reward verifying the initial term and substituting the formula into the recurrence to recover $u_{n+1}$ .