Sequences: Formulas and Recurrence

Generate terms from an nth-term formula or a recurrence relation, and decide whether a sequence is increasing, decreasing or periodic.

In the exam for Edexcel International AS Maths · 9 real past-paper questions below

Two ways to make a sequence

A sequence is a list of numbers made by a rule. There are two kinds of rule. One jumps straight to any position. The other builds each term from the one before.

Formula for the nth term. Put in the position $n$ and get the term. Example: $u_{n} = 4 n - (- 1)^{n}$ .

Recurrence relation. Use the term you have to make the next one. Example: $x_{n + 1} = 2 x_{n} + 3$ , with $x_{1} = 1$ .

Formula

x_{n + 1} = f (x_{n}), x_{1} = a given number

$x_{n}$ — the term you already have.
$x_{n + 1}$ — the next term, made from it.
$x_{1}$ — the starting term. Without it the recurrence cannot start.

Worked example

1
Find the first four terms of $x_{n + 1} = 2 x_{n} + 3$ , with $x_{1} = 1$ .
2
$x_{2} = 2 (1) + 3 = 5$ — Put $n = 1$ : the rule uses $x_{1}$ to make $x_{2}$ .
3
$x_{3} = 2 (5) + 3 = 13$ — Put $n = 2$ : use $x_{2}$ .
4
$x_{4} = 2 (13) + 3 = 29$ — Put $n = 3$ : use $x_{3}$ . Each term needs the one before it.

Worked example

1
For $u_{n} = 4 n - (- 1)^{n}$ , find $u_{3}$ and $u_{4}$ .
2
$u_{3} = 12 - (- 1)^{3} = 12 + 1 = 13$ — Odd $n$ : $(- 1)^{n} = - 1$ .
3
$u_{4} = 16 - (- 1)^{4} = 16 - 1 = 15$ — Even $n$ : $(- 1)^{n} = 1$ . No earlier terms needed.

Common mistake

Reading $x_{n + 1}$ as $x_{n} + 1$ . The $n + 1$ is a position label, the term after $x_{n}$ . It is not “add 1”.

Key idea

Fibonacci described a recurrence in 1202, about breeding rabbits: each term is the sum of the two before, 1, 1, 2, 3, 5, 8, … Banks use the same idea today: next year’s balance = this year’s balance × 1.05.

Key idea

A formula jumps to any term. A recurrence walks there one step at a time, from the first term.

Recurrences with an unknown

Exam questions often put a letter, such as $k$ , in the recurrence. You find terms in terms of $k$ , then a fact about the terms gives an equation for $k$ .

Worked example

1
A sequence has $u_{1} = 4$ and $u_{n + 1} = k u_{n} - 2$ . Find $u_{2}$ and $u_{3}$ in terms of $k$ .
2
$u_{2} = k (4) - 2 = 4 k - 2$ — Put $u_{1} = 4$ into the rule.
3
$u_{3} = k (4 k - 2) - 2$ — Put the whole of $u_{2}$ in, inside a bracket.
4
$u_{3} = 4 k^{2} - 2 k - 2$ — Expand. This is a quadratic in $k$ .

Worked example

1
Given that $\sum_{n = 1}^{3} u_{n} = 42$ and $k > 0$ , find $k$ and $u_{2}$ .
2
$4 + (4 k - 2) + (4 k^{2} - 2 k - 2) = 42$ — The sum of the first three terms. Write all three.
3
$4 k^{2} + 2 k - 42 = 0$ — Collect terms.
4
$2 k^{2} + k - 21 = 0$ — Divide by 2.
5
$(k - 3) (2 k + 7) = 0$ — Factorise.
6
$k = 3 or k = - \frac{7}{2}$ — Two roots.
7
$k = 3, u_{2} = 4 (3) - 2 = 10$ — $k > 0$ rejects $- \frac{7}{2}$ . Then use your $u_{2}$ expression.

Worked example

1
Same sequence, but now you are told $u_{3} = 10$ . Find the possible values of $k$ .
2
$4 k^{2} - 2 k - 2 = 10$ — Set your $u_{3}$ expression equal to the given value.
3
$2 k^{2} - k - 6 = 0 \Rightarrow (k - 2) (2 k + 3) = 0$ — Tidy up and factorise.
4
$k = 2 or k = - \frac{3}{2}$ — Keep both unless the question rules one out.

Marks: M1 for using the rule to build the terms, M1 for adding them and setting the sum equal to the total, M1 for solving the quadratic, A1 for the right value. If the question says “show that” the quadratic is …, every line must be seen.

Common mistake

Dropping the bracket: writing $k \times 4 k - 2 - 2$ instead of $k (4 k - 2) - 2$ . Always put the whole previous term in a bracket.

Common mistake

Adding only $u_{2} + u_{3}$ . $\sum_{n = 1}^{3}$ starts at $n = 1$ , so $u_{1}$ is in the sum too.

Key idea

Build the terms in $k$ , turn the given fact into an equation, solve, then use any condition to choose.

Increasing, decreasing or periodic

Questions ask you to describe how a sequence behaves. There are three words to know.

Increasing: every term is bigger than the one before, $u_{n + 1} > u_{n}$ for all $n$ . Example: 1, 5, 13, 29, …
Decreasing: every term is smaller than the one before, $u_{n + 1} < u_{n}$ for all $n$ . Example: 20, 10, 5, 2.5, …
Periodic: the terms repeat in a cycle. The order (or period) is the number of terms in one cycle.

Worked example

1
Show that $u_{1} = 2$ , $u_{n + 1} = 1 - \frac{1}{u _{n}}$ is periodic, and state its order.
2
$u_{2} = 1 - \frac{1}{2} = \frac{1}{2}$ — Apply the rule to $u_{1}$ .
3
$u_{3} = 1 - \frac{1}{\frac{1}{2}} = 1 - 2 = - 1$ — Dividing by a half doubles.
4
$u_{4} = 1 - \frac{1}{- 1} = 1 + 1 = 2$ — Back to the start.
5
$u_{4} = u_{1}$ — Say it: $u_{4} = u_{1}$ , so the terms repeat. The sequence is periodic.
6
$order 3$ — Three terms in each cycle: 2, ½, −1.

A sequence that switches sign with $(- 1)^{n}$ can also be periodic. Here $u_{1} = 3$ and $u_{n + 1} = (- 1)^{n + 1} u_{n} + 5$ gives 3, 8, −3, 2, 3, … which has order 4.

Common mistake

Listing the terms and stopping. For “show that it is periodic”, you must also state that a term equals $u_{1}$ , so it repeats.

Common mistake

Writing “it repeats every third term” for the order. The mark scheme wants the number: order 3.

Key idea

Work out terms until one matches $u_{1}$ . The number of steps it took is the order.

Summing a periodic sequence

Exams often end with a long sum. You never add 80 terms one by one. A periodic sequence is made of identical blocks, so you add one block and multiply.

Formula

n = 1 \sum N u_{n} = q \times S_{cycle} + leftovers

$S_{cycle}$ — the sum of one full cycle.
$q$ — the number of full cycles: divide $N$ by the order $p$ and take the whole number part.
$leftovers$ — the remainder $r$ terms. They are the first $r$ terms of the cycle, $u_{1}$ to $u_{r}$ .

Steps

1
Cycle: list one cycle and add it.
2
Count: $N \div p = q$ remainder $r$ .
3
Leftovers: add $u_{1}, \dots, u_{r}$ , then total $q \times S_{cycle} + leftovers$ .

Worked example

1
Use $u_{1} = 2, u_{n + 1} = 1 - \frac{1}{u _{n}}$ (cycle 2, ½, −1) to find $\sum_{n = 1}^{80} u_{n}$ .
2
$S_{cycle} = 2 + \frac{1}{2} + (- 1) = \frac{3}{2}$ — Cycle: add one full cycle.
3
$80 = 26 \times 3 + 2$ — Count: 26 full cycles, 2 terms left over.
4
$26 \times \frac{3}{2} = 39$ — The full cycles.
5
$u_{1} + u_{2} = 2 + \frac{1}{2} = \frac{5}{2}$ — Leftovers: the first 2 terms of the cycle.
6
$39 + \frac{5}{2} = \frac{83}{2} = 41.5$ — Add.

The same count finds a far term. For $u_{100}$ : $100 = 33 \times 3 + 1$ , so $u_{100}$ is the 1st term of a cycle, $u_{100} = u_{1} = 2$ . If the remainder is 0, it is the last term of the cycle.

Memory trick

It is Monday. What day is it in 100 days? 100 = 14 weeks + 2 days, so Wednesday. Weekdays are a periodic sequence of order 7: whole weeks change nothing, only the leftover days count.

Worked example

1
$u_{1} = 3$ , $u_{n + 1} = (- 1)^{n + 1} u_{n} + k$ . Given $\sum_{n = 1}^{4} u_{n} = 10$ , find $k$ , then $\sum_{n = 1}^{50} u_{n}$ .
2
$u_{2} = 3 + k, u_{3} = - 3, u_{4} = k - 3, u_{5} = 3$ — Use the rule four times. Watch the sign of $(- 1)^{n + 1}$ each time.
3
$3 + (3 + k) - 3 + (k - 3) = 2 k = 10$ — The sum of one cycle.
4
$k = 5 : 3, 8, - 3, 2$ — The cycle.
5
$50 = 12 \times 4 + 2$ — 12 full cycles, 2 left over.
6
$12 \times 10 + (3 + 8) = 131$ — Full cycles plus $u_{1} + u_{2}$ .

Marks: M1 for a correct method (cycles × cycle sum + leftovers, or the numbers written out in that shape), A1 for the answer.

Common mistake

Using the AP or GP sum formula. A periodic sequence is neither, and that method scores no marks.

Common mistake

Forgetting the leftovers, or taking them from the end of the cycle. The leftovers always start again at $u_{1}$ .

Key idea

Cycle, count, leftovers: one cycle’s sum × the number of full cycles, plus the first few terms again.

A formula with a repeating part

Some nth-term formulas mix a normal part with a part that repeats, such as $(- 1)^{n}$ . Split the sum into two sums and do each one with its own tool.

Worked example

1
$u_{n} = 3 n + (- 1)^{n}$ . Find $\sum_{n = 1}^{41} u_{n}$ .
2
$\sum_{n = 1}^{41} 3 n + \sum_{n = 1}^{41} (- 1)^{n}$ — Split the sum into two.
3
$3 + 6 + 9 + \dots + 123$ — The first sum is an AP: $a = 3$ , $d = 3$ , 41 terms.
4
$\frac{41}{2} (3 + 123) = 2583$ — AP sum, $\frac{n}{2} (a + l)$ .
5
$(- 1) + 1 + (- 1) + \dots$ — The second is periodic, order 2, cycle sum 0.
6
$41 = 20 \times 2 + 1 \Rightarrow 20 \times 0 + (- 1) = - 1$ — 20 full cycles add 0. One leftover: $u_{1}$ ’s part, −1.
7
$2583 + (- 1) = 2582$ — Add the two sums.

Trig formulas repeat too. $u_{n} = cos (\frac{nπ}{2})$ gives 0, −1, 0, 1, 0, … which is periodic with order 4 and cycle sum 0. Work in radians and find the first few terms exactly, then use cycle, count, leftovers.

Marks: M1 for the AP sum with the right $a$ , $d$ and number of terms, then A1. M1 A1 again for the repeating part. Show both sums: a bare answer can score nothing. Writing it out as odd and even terms also works.

Common mistake

Treating the whole thing as one AP. It is not: the gaps go 3 + 2 = 5, then 3 − 2 = 1, and so on.

Key idea

Normal part: AP or GP formula. Repeating part: cycle, count, leftovers. Then add.

Which method? Read the question

The wording tells you which step to use.

You see	You do
$u_{n} =$ a formula in $n$	Put the number $n$ straight in
$u_{n + 1} =$ a rule with $u_{n}$	Start from $u_{1}$ , one term at a time
“Find $u_{3}$ in terms of $k$ ”	Put the whole previous term in a bracket; expand
A sum or a term is given as a number	Set your expression equal to it; solve, often a quadratic
“Show that the sequence is periodic”	Find terms until one equals $u_{1}$ ; say so
“State the order”	The number of terms in one cycle
$\sum$ of a periodic sequence, or a far term	Cycle, count, leftovers
A formula with $(- 1)^{n}$ added	Split: AP part + repeating part

Common mistake

Using a calculator list with no working. On “show that” questions, every term must be written out.

Key idea

Write each term clearly, keep brackets, and for long sums let the cycle do the work.

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