4.3 Sequences from formulae and recurrences — Practise

Work through these on your own. Decide first whether the question gives a formula for the $n$ th term or a recurrence. A formula reaches any term directly; a recurrence has to travel through every term on the way.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A sequence is given by $u_n = 4n - 7$ . Work out $u_{20}$ .
A sequence is given by $u_n = 3n^2 + 1$ . Work out $u_6$ .
A sequence is given by $u_{n+1} = 2u_n + 3$ and $u_1 = 1$ . Work out $u_4$ .
A sequence is given by $u_{n+1} = u_n - 4$ and $u_1 = 20$ . Find the first term that is negative.
A sequence repeats the pattern $6, 1, 8, 1, 6, 1, 8, 1, \dots$ . Work out $u_{50}$ .
The Fibonacci sequence is defined by $f_1 = 1$ , $f_2 = 1$ and $f_{n+2} = f_{n+1} + f_n$ . Work out $f_{10}$ .
A sequence is given by $u_n = (-1)^n$ . Work out $u_7$ and $u_{100}$ , and state the period.
A van costs £12,000. Its value falls by 15% each year, so $v_{n+1} = 0.85v_n$ with $v_1 = 12{,}000$ . Work out its value after 3 years, to the nearest penny.
A sequence is given by $u_{n+1} = \frac{1}{2}u_n + 1$ and $u_1 = 0$ . Work out $u_4$ and say whether the sequence increases or decreases.
The sequence $1, 4, 9, 16, \dots$ can be described by the formula $u_n = n^2$ or by the recurrence $u_{n+1} = u_n + 2n + 1$ . Show that the recurrence gives the gap $7$ between the third and fourth terms, then work out $u_{20}$ .
A sequence is given by $u_{n+1} = 2u_n - 1$ and $u_1 = 3$ . Work out $u_5$ .
The quiz night follows $q_{n+1} = q_n + 50$ with $q_1 = 150$ . The online appeal follows $a_{n+1} = 0.8a_n$ with $a_1 = 600$ . Work out the quiz income in month 10 and the appeal income in month 6, to the nearest penny.

Answers and concise solutions

1. The term

Answer: Substitute $n = 20$ : $u_{20} = 4(20) - 7 = 80 - 7 = 73$ . The formula reaches position $20$ directly, so the eighteen terms in between are not needed.

2. The term

Answer: Substitute $n = 6$ : $u_6 = 3(36) + 1 = 108 + 1 = 109$ . The square comes before the multiplication, so $3 \times 6^2$ is $3 \times 36$ , not $(3 \times 6)^2$ .

3. The term

Answer: Work forward one term at a time: $u_2 = 2(1) + 3 = 5$ , then $u_3 = 2(5) + 3 = 13$ , then $u_4 = 2(13) + 3 = 29$ . The starting value $u_1 = 1$ is part of the definition; without it the rule gives no sequence.

4. The position of the first negative term

Answer: The terms fall by $4$ each step: $20, 16, 12, 8, 4, 0, -4, \dots$ The sixth term is exactly $0$ , which is not negative. The seventh term is $-4$ , so the first negative term is $u_7 = -4$ . The rule for the $n$ th term is $u_n = 20 - 4(n - 1) = 24 - 4n$ .

5. The term

Answer: The pattern repeats every $4$ terms, so the period is $4$ . Now $50 = 4 \times 12 + 2$ , so the remainder is $2$ . Position $50$ is therefore the second term of a block, which is $1$ . So $u_{50} = 1$ . The remainder selects the term; a remainder of $0$ would mean the fourth term of a block.

6. The term

Answer: Each term is the sum of the two before it: $1, 1, 2, 3, 5, 8, 13, 21, 34, 55$ . So $f_{10} = 55$ . This recurrence needs two starting values, because the rule adds the two previous terms.

7. The two terms and the period

Answer: An odd power of $-1$ is $-1$ , so $u_7 = -1$ ; an even power is $1$ , so $u_{100} = 1$ . The terms are $-1, 1, -1, 1, \dots$ , so the sequence is periodic with period $2$ .

8. The value

Answer: The value after 3 years is $v_4 = 12{,}000(0.85)^3 = 12{,}000 \times 0.614125 = 7369.50$ . The factor $0.85$ appears once for each year, so it is cubed after three years. Losing 15% each year is a decreasing sequence that settles towards zero.

9. The term and the direction

Answer: The terms are $u_1 = 0$ , $u_2 = 1$ , $u_3 = \frac{3}{2}$ and $u_4 = \frac{7}{4}$ . Each term is larger than the one before, so the sequence increases. The steps get smaller, and the terms settle towards $2$ , which is the value that satisfies $u = \frac{1}{2}u + 1$ .

10. The gap and the twentieth term

Answer: The gap between $u_3 = 9$ and $u_4 = 16$ is $16 - 9 = 7$ , and the recurrence gives $2(3) + 1 = 7$ at $n = 3$ . Both descriptions fit. The formula is better for a distant term: $u_{20} = 20^2 = 400$ , which the recurrence would reach only after nineteen steps. This is the choice between a formula and a recurrence in practice.

11. The term

Answer: Work forward: $u_2 = 5$ , $u_3 = 9$ , $u_4 = 17$ , $u_5 = 33$ . Four steps are needed, because the rule moves from one term to the next. The sequence increases, and the steps roughly double each time.

12. The two incomes

Answer: The quiz income in month 10 is $q_{10} = 150 + 9(50) = 600$ , so £600. The appeal income in month 6 is $a_6 = 600(0.8)^5 = 600 \times 0.32768 = 196.608$ , so £196.61 to the nearest penny. The quiz stream grows without limit, while the appeal stream shrinks towards zero.