Recurrence rules and explicit terms | Year 11

01 / LEARN

The idea, step by step

Two ways to describe a list

A sequence is an ordered list. The index identifies a position: $u_{1}$ normally means the first term, whereas $u_{0}$ means the starting state in an index-zero model. An explicit rule calculates a selected term directly from its index. A recurrence calculates a new term from earlier terms.

A recurrence also needs starting values. For example, $u_{n + 1} = u_{n} + 3$ permits many sequences until an initial term is supplied. Record the result of one step before beginning the next step.

u_{1} = 4, u_{n + 1} = u_{n} + 3 ⟺ u_{n} = 3 n + 1

Checking and interpreting a rule

Substitute several allowed indices to check a proposed rule. Agreement with a few terms suggests a possible pattern but does not prove it is the only possible rule. An explicit rule is convenient for a distant term; a recurrence can better describe a repeating process.

State the domain of the index. A physical count may allow only whole-number times, even if the formula can be evaluated at decimals or negative numbers. Do not interpret such values unless the model supports them.

A common trap

A recurrence by itself determines the sequence.

Instead: An initial value is also required; changing it can change every term.

02 / WORKED EXAMPLES

See the thinking

Example 1

A distant term

For $u_{n} = 5 n - 2$ , $n \geq 1$ , find $u_{1}, u_{4}, u_{20}$ .

$u_{1} = 5 (1) - 2 = 3$ .
$u_{4} = 5 (4) - 2 = 18$ and $u_{20} = 5 (20) - 2 = 98$ .

3, 18, 98

Example 2

A nonlinear recurrence

$v_{0} = 2$ , $v_{n + 1} = v_{n}^{2} - 1$ . Find $v_{1}, v_{2}, v_{3}$ .

$v_{1} = 2^{2} - 1 = 3$ .
$v_{2} = 3^{2} - 1 = 8$ ; $v_{3} = 8^{2} - 1 = 63$ . The starting index is zero.

3, 8, 63

Check your understanding

$a_{1} = 6$ , $a_{n + 1} = a_{n} - 2$ . Find $a_{4}$ .

Show answer

$0$ .

The terms are $6, 4, 2, 0$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Find the first four terms of $u_{n} = 2 n^{2} - 1$ , $n \geq 1$ .
2 marks
$a_{1} = 5$ , $a_{n + 1} = 2 a_{n} - 1$ . Find $a_{2}, a_{3}, a_{4}$ .
2 marks
For $b_{n} = 30 - 4 n$ , find $b_{7}$ . Can a term equal zero for integer $n \geq 1$ ?
2 marks
$c_{0} = 10$ , $c_{n + 1} = 0.5 c_{n}$ . Find $c_{3}$ .
2 marks
Give an explicit rule for $7, 11, 15, 19, \dots$ with first index 1.
2 marks
Give two different sequences satisfying $u_{n + 1} = u_{n} + 5$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find the first four terms of $u_{n} = 2 n^{2} - 1$ , $n \geq 1$ .
2 marks
Substitute $n = 1, 2, 3, 4$ separately.
$1, 7, 17, 31$ .
$a_{1} = 5$ , $a_{n + 1} = 2 a_{n} - 1$ . Find $a_{2}, a_{3}, a_{4}$ .
2 marks
$2 (5) - 1 = 9$ , then $2 (9) - 1 = 17$ , then $2 (17) - 1 = 33$ .
$9, 17, 33$ .
For $b_{n} = 30 - 4 n$ , find $b_{7}$ . Can a term equal zero for integer $n \geq 1$ ?
2 marks
$30 - 28 = 2$ . The equation $30 - 4 n = 0$ gives $n = 7.5$ , not an allowed index.
$b_{7} = 2$ ; no zero term.
$c_{0} = 10$ , $c_{n + 1} = 0.5 c_{n}$ . Find $c_{3}$ .
2 marks
$c_{1} = 5, c_{2} = 2.5, c_{3} = 1.25$ .
$1.25$ .
Give an explicit rule for $7, 11, 15, 19, \dots$ with first index 1.
2 marks
$7 + 4 (n - 1) = 4 n + 3$ .
$u_{n} = 4 n + 3$ .
Give two different sequences satisfying $u_{n + 1} = u_{n} + 5$ .
3 marks
Different initial values satisfy the same recurrence.
For example $1, 6, 11, \dots$ and $3, 8, 13, \dots$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics sequence and series content with introductory induction, regrouped into an original school-sized chapter.

Official curriculum source ↗

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