Recurrence rules and explicit terms | Year 11

01 / LEARN

Understand the idea.

Two ways to describe a list

An explicit rule takes an index as its input and returns the term at that position. Read the starting index first: $u_{1}$ is the first term in an index-one sequence, while $u_{0}$ is the initial state in an index-zero model.

A recurrence uses earlier terms to generate later ones. It needs an initial value as well as an update rule. For $u_{n + 1} = u_{n} + 3$ , choosing $u_{1} = 4$ fixes the sequence; choosing a different starting value changes every term.

\begin{array}{c} u_{1} = 4, u_{n + 1} = u_{n} + 3 \\ ⟺ u_{n} = 3 n + 1 \end{array}

Checking and interpreting a rule

To evaluate an explicit rule, replace every occurrence of $n$ with the requested index. To use a recurrence, finish one update and feed that result into the next. The subscript labels a position; it is not a multiplier.

A few matching terms can disprove a wrong proposed rule, but cannot establish that a pattern continues forever or is unique. Also respect the index domain: a formula may accept $n = 2.5$ algebraically even though a sequence has no term at that position.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A distant term

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

Worked solution

1
Substitute the first index.
$u_{1} = 5 (1) - 2 = 3$
The first term uses the allowed index one.
2
Evaluate the fourth term directly.
$u_{4} = 5 (4) - 2 = 18$
An explicit rule does not require the intervening terms.
3
Evaluate the twentieth term.
$u_{20} = 5 (20) - 2 = 98$
The same formula applies at every permitted integer index.

Answer

$3, 18, 98$ .

EXAMPLE 02

A nonlinear recurrence

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

Worked solution

1
Use the given starting state.
$v_{1} = v_{0}^{2} - 1 = 2^{2} - 1 = 3$
The update from index zero produces the term at index one.
2
Feed the new value into the recurrence.
$v_{2} = v_{1}^{2} - 1 = 3^{2} - 1 = 8$
Each square uses the immediately preceding term.
3
Update once more.
$v_{3} = v_{2}^{2} - 1 = 8^{2} - 1 = 63$
Three updates are needed to move from index zero to index three.

Answer

$3, 8, 63$ .

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$0$ .

Starting at $a_{1} = 6$ , subtract two three times: $a_{2} = 4$ , $a_{3} = 2$ , $a_{4} = 0$ . The fourth term is three updates after the first.

Keep practising

6 worksheet questions · 13 marks

Open worksheet 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 ↗

In this lesson

Understand the idea.

Two ways to describe a list

Checking and interpreting a rule

See the method in action.

A distant term

Substitute the first index.

Evaluate the fourth term directly.

Evaluate the twentieth term.

A nonlinear recurrence

Use the given starting state.

Feed the new value into the recurrence.

Update once more.

Try it yourself.

Check your understanding

Keep practising