4.3 Sequences from formulae and recurrences — Practise with help

Work each one first, closed book. When you meet a recurrence, write the starting value down beside it before you substitute.

Warm-up: remember

Work out $2(4)^2 - 3$ .
Work out $3 \times 5 + 2$ and $3(5 + 2)$ .
Write down the next two terms of $2, 5, 8, 11, \dots$ .

Answers

$2(16) - 3 = 32 - 3 = 29$ .
$3 \times 5 + 2 = 17$ , and $3(5 + 2) = 3 \times 7 = 21$ . The bracket changes the answer.
The terms rise by $3$ , so the next two are $14$ and $17$ .

If question 1 was hard, revisit the order of operations before you continue. The power comes before the multiplication.

Practise with help

1. A sequence is given by $u_n = 3n + 1$ . Write down the first four terms.

Hint

The subscript is the position. The first term is $u_1$ , so substitute $n = 1$ into the rule.

Next step

Substitute $n = 1, 2, 3, 4$ in turn. The first term is $u_1 = 3(1) + 1$ .

Full working

u_1 = 3(1) + 1 = 4, \qquad u_2 = 3(2) + 1 = 7,

u_3 = 3(3) + 1 = 10, \qquad u_4 = 3(4) + 1 = 13.

The first four terms are $4, 7, 10, 13$ . The sequence is increasing.

2. A sequence is given by $u_n = n^2 - 2n$ . Work out $u_5$ and $u_{10}$ .

Hint

The rule gives any term from its position, so there is no need to work out the terms in between.

Next step

Substitute $n = 5$ : the first term of the expression is $5^2 = 25$ , and the second is $2 \times 5 = 10$ . Then substitute $n = 10$ .

Full working

u_5 = 5^2 - 2(5) = 25 - 10 = 15,

u_{10} = 10^2 - 2(10) = 100 - 20 = 80.

The tenth term is $80$ . Notice how much faster the terms grow than in question 1, because the rule contains $n^2$ .

3. A sequence is given by $u_{n+1} = 2u_n + 1$ and $u_1 = 3$ . Work out $u_2$ , $u_3$ and $u_4$ .

Hint

Each step uses the term you found a moment ago. Put the value into the rule to get the next one.

Next step

Start with $u_1 = 3$ , so $u_2 = 2(3) + 1$ . Then put the value of $u_2$ back into the same rule.

Full working

u_2 = 2u_1 + 1 = 2(3) + 1 = 7,

u_3 = 2u_2 + 1 = 2(7) + 1 = 15,

u_4 = 2u_3 + 1 = 2(15) + 1 = 31.

The first four terms are $3, 7, 15, 31$ . Each term is larger than the one before.

4. A sequence is given by $u_{n+1} = 5 - u_n$ and $u_1 = 1$ . Write down the first five terms and describe the sequence.

Hint

Work out one term at a time. Watch what happens when a term is bigger than $5$ , because the rule subtracts it from $5$ .

Next step

$u_2 = 5 - 1 = 4$ . Then $u_3 = 5 - 4 = 1$ . Carry on for two more terms and look for a pattern.

Full working

u_1 = 1, \quad u_2 = 4, \quad u_3 = 1, \quad u_4 = 4, \quad u_5 = 1.

The terms repeat in the pattern $1, 4$ . The sequence is periodic with period $2$ , so $u_{n+2} = u_n$ for every $n$ . It is neither increasing nor decreasing.

5. A sequence repeats the pattern $4, 9, 4, 9, \dots$ . Work out $u_{50}$ .

Hint

Find the period first. Then divide the position by the period and look at the remainder.

Next step

The period is $2$ . Work out $50 \div 2$ and find the remainder, then read off the matching term of the block.

Full working

The period is $2$ , and $50 = 2 \times 25 + 0$ . A remainder of $0$ means the last term of a block, which is $9$ . So $u_{50} = 9$ . The even positions all hold $9$ and the odd positions all hold $4$ .

6. The quiz night in the chapter problem raises £150 in month 1 and £50 more each month. Its income follows $q_{n+1} = q_n + 50$ with $q_1 = 150$ . Work out the income in month 6.

Hint

Month 6 is five steps after month 1, so there are five additions of £50. You can count forward or use the pattern.

Next step

Month 6 is $q_6 = q_1 + 5 \times 50 = 150 + 250$ .

Full working

q_6 = q_1 + 5(50) = 150 + 250 = 400.

The quiz night raises £400 in month 6. The income in month $n$ is $q_n = 150 + 50(n - 1)$ , so the sequence increases by £50 each month.

7. The online appeal raises £600 in month 1 and 80% of that each month after. Its income follows $a_{n+1} = 0.8a_n$ with $a_1 = 600$ . Work out the income in month 4, to the nearest penny.

Hint

Month 4 is three steps after month 1, so the rule is applied three times. Each step multiplies by $0.8$ .

Next step

$a_4 = 600 \times 0.8 \times 0.8 \times 0.8 = 600 \times 0.8^3$ . Work out $0.8^3 = 0.512$ first.

Full working

a_4 = 600(0.8)^3 = 600 \times 0.512 = 307.2.

The appeal income in month 4 is £307.20. The sequence decreases, and each term is 80% of the one before. The general rule is $a_n = 600(0.8)^{n-1}$ .