2.1 Sequences, Recurrence Relations and Series
SL 1.2
- the words sequence, term, general term, finite and infinite
- how to generate a sequence from a general term \(u_n\) or from a recurrence relation
- how to make a conjecture for the general term from the first few terms
- what a series is, and the notation \(S_n\)
- how to find \(u_n\) from \(S_n\) using \(u_n = S_n - S_{n-1}\)
Sequences
A sequence is a list of numbers written in a definite order. Each number in the list is a term.
\[ \text{(a)}\;\; 2,\, 5,\, 8,\, 11,\, \ldots,\, 29 \qquad\qquad \text{(b)}\;\; 1,\, 3,\, 9,\, 27,\, 81,\, \ldots \]
Sequence (a) stops at \(29\): it is a finite sequence. Sequence (b) goes on for ever: it is an infinite sequence.
We name the terms with a letter and a position number (a subscript):
\[ u_1,\; u_2,\; u_3,\; \ldots,\; u_n,\; \ldots \]
\(u_1\) is the first term, \(u_2\) is the second term, and \(u_n\) is the \(n\)th term. \(u_n\) is also called the general term, because it describes every term at once. The whole sequence is sometimes written \(\{u_n\}\).
Generating sequences from the general term
If you know a formula for \(u_n\), substitute \(n = 1, 2, 3, \ldots\) to find the terms.
Example 1 Write down the first three terms of the sequence with general term \(u_n = 2n^2 + 1\), where \(n = 1, 2, 3, \ldots\)
\[ u_1 = 2(1)^2 + 1 = 3, \qquad u_2 = 2(2)^2 + 1 = 9, \qquad u_3 = 2(3)^2 + 1 = 19 \]
The first three terms are \(3\), \(9\), \(19\).
Recurrence relations
Sometimes each term is made from the term before it. Look at this sequence:
| Term | Rule | Value | |
|---|---|---|---|
| first term | \(u_1\) | \(4\) | |
| second term | \(u_2\) | \(u_1 + 3\) | \(7\) |
| third term | \(u_3\) | \(u_2 + 3\) | \(10\) |
| fourth term | \(u_4\) | \(u_3 + 3\) | \(13\) |
| \(\vdots\) | |||
| \((n+1)\)th term | \(u_{n+1}\) | \(u_n + 3\) |
We can describe the whole sequence in one line:
\[ u_1 = 4, \qquad u_{n+1} = u_n + 3, \qquad n \in \mathbb{Z}^+ . \]
This is a recurrence relation (also called a recursive formula). It needs two parts: a starting value and a rule that makes the next term from the previous one.
Example 2 A sequence is defined by \(u_1 = 2\) and \(u_{n+1} = 2u_n - 1\), for \(n \in \mathbb{Z}^+\). Write down the first three terms.
\[ u_1 = 2, \qquad u_2 = 2(2) - 1 = 3, \qquad u_3 = 2(3) - 1 = 5 \]
Recurrence relations are natural in real life, because many situations are described “from one year to the next”.
Example 3 A pond has 500 fish at the start of the first year. Each year the number of fish increases by 20%, and then 50 fish are removed at the end of the year. Let \(u_n\) be the number of fish at the start of year \(n\).
(a) Write down a recurrence relation for \(u_n\).
(b) Find the number of fish at the start of the third year.
(a) An increase of 20% multiplies the number by \(1.2\). Then 50 are removed:
\[ u_1 = 500, \qquad u_{n+1} = 1.2\,u_n - 50 . \]
(b) \(u_2 = 1.2(500) - 50 = 550\) and \(u_3 = 1.2(550) - 50 = 610\).
There are 610 fish at the start of the third year.
Lists & Spreadsheet page: Menu → Data → Generate Sequence. For the fish in Example 3, type the formula u(n)=1.2·u(n-1)-50, Initial Terms 500, n0 1, nMax 10. The terms appear in a column.
Graphs page: Menu → Graph Entry/Edit → Sequence → Sequence. Type u1(n)=1.2·u1(n-1)-50 with Initial Terms 500. Then ctrl + T (split-screen table) shows the values.
Conjectures
Look at the numbers \(u_1 = 1\), \(u_2 = 3\), \(u_3 = 5\), \(u_4 = 7\). Each term is \(2\) more than the one before. A formula that fits all four terms is
\[ u_n = 2n - 1 . \]
A statement like this is a conjecture: it appears to be true for every \(n\), but we have not proved it. (You will learn how to prove such statements in Unit 6.)
Example 4 Make a conjecture for the general term \(u_n\) of the sequence
\[ \frac{2}{3},\; \frac{4}{5},\; \frac{6}{7},\; \frac{8}{9},\; \ldots \]
Look at the numerators and the denominators separately.
- Numerators: \(2, 4, 6, 8, \ldots\) — these are \(2n\).
- Denominators: \(3, 5, 7, 9, \ldots\) — these are one more than the numerators, \(2n + 1\).
Conjecture: \(u_n = \dfrac{2n}{2n + 1}\).
Example 5 The sequence \(\{u_n\}\) is defined by \(u_1 = 3\) and \(u_{n+1} = u_n + 4\). Make a conjecture for a formula for \(u_n\) in terms of \(n\).
The terms are \(3, 7, 11, 15, \ldots\) We add \(4\) each time.
\[ u_1 = 3, \quad u_2 = 3 + 4, \quad u_3 = 3 + 2 \times 4, \quad u_4 = 3 + 3 \times 4 \]
The number of 4s is one less than \(n\), so \(u_n = 3 + 4(n - 1) = 4n - 1\).
Series
When we add the terms of a sequence, we get a series. A finite series has a finite number of terms; an infinite series goes on for ever.
The sum of the first \(n\) terms is written \(S_n\):
\[ S_n = u_1 + u_2 + u_3 + \cdots + u_{n-1} + u_n . \]
Since \(S_n = S_{n-1} + u_n\), we can find any term from the sums:
The \(n\)th term from the sum Not in the formula booklet
\[ u_n = S_n - S_{n-1}, \qquad n \ge 2 . \]
For the first term, \(u_1 = S_1\).
- \(u_n\): the \(n\)th term
- \(S_n\): the sum of the first \(n\) terms
- \(S_{n-1}\): the sum of the first \(n - 1\) terms
- \(n\): the position of the term
Example 6 The sum of the first \(n\) terms of a series is \(S_n = 3^n - 1\). Find
(a) the \(n\)th term of the series,
(b) the first and second terms.
(a) For \(n \ge 2\):
\[ u_n = S_n - S_{n-1} = \left(3^n - 1\right) - \left(3^{n-1} - 1\right) = 3^n - 3^{n-1} = 3^{n-1}(3 - 1) = 2 \times 3^{n-1} . \]
Check \(n = 1\): \(S_1 = 3 - 1 = 2\) and \(2 \times 3^0 = 2\) ✓, so the formula works for every \(n\).
(b) \(u_1 = 2\) and \(u_2 = 2 \times 3 = 6\). (Check: \(S_2 = 9 - 1 = 8 = 2 + 6\) ✓)
Example 7 The sum of the first \(n\) terms of a series is \(S_n = 2n^2 + n\), where \(n \in \mathbb{Z}^+\).
(a) Find the \((n + 1)\)th term, \(u_{n+1}\).
(b) By considering \(S_n - S_{n-1}\), show that \(u_n = 4n - 1\), and check that this agrees with (a).
(a) \(u_{n+1} = S_{n+1} - S_n\):
\[ \left(2(n+1)^2 + (n+1)\right) - \left(2n^2 + n\right) = 2n^2 + 4n + 2 + n + 1 - 2n^2 - n = 4n + 3 . \]
(b)
\[ S_n - S_{n-1} = \left(2n^2 + n\right) - \left(2(n-1)^2 + (n-1)\right) = 2n^2 + n - \left(2n^2 - 4n + 2 + n - 1\right) = 4n - 1 . \]
Replacing \(n\) by \(n + 1\) in \(4n - 1\) gives \(4(n + 1) - 1 = 4n + 3\), which agrees with (a). ✓
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(u_{n+1} = u_n + 1\) | \(u_{n+1}\) is the next term | The subscript is a position, not a value. |
| Recurrence \(u_{n+1} = 2u_n - 1\) with no \(u_1\) | \(u_1 = 2\), \(u_{n+1} = 2u_n - 1\) | A recurrence relation needs a starting value. |
| A conjecture is proved by checking 4 terms | It is only a guess that fits | Proof needs an argument for every \(n\). |
| \(S_n\) is the \(n\)th term | \(S_n\) is the sum of the first \(n\) terms | \(u_n\) is a term; \(S_n\) is a sum. |
| \((n-1)^2 = n^2 - 1\) | \((n-1)^2 = n^2 - 2n + 1\) | Expand the bracket fully when you find \(S_{n-1}\). |
Exercises
1 Write down the first four terms of the sequence \(u_n = n^2 - 3n\).
\(u_1 = -2\), \(u_2 = -2\), \(u_3 = 0\), \(u_4 = 4\).
2 Write down the first three terms of the sequence \(u_n = (-1)^n\,2^n\).
\(u_1 = -2\), \(u_2 = 4\), \(u_3 = -8\). The factor \((-1)^n\) makes the signs alternate.
3 A sequence is defined by \(u_1 = 5\) and \(u_{n+1} = 3 - 2u_n\). Find \(u_2\), \(u_3\) and \(u_4\).
\(u_2 = 3 - 10 = -7\), \(\quad u_3 = 3 + 14 = 17\), \(\quad u_4 = 3 - 34 = -31\).
4 A sequence is defined by \(u_1 = 1\), \(u_2 = 1\) and \(u_{n+2} = u_{n+1} + 2u_n\). Find the first five terms.
\(u_3 = 1 + 2(1) = 3\), \(\;u_4 = 3 + 2(1) = 5\), \(\;u_5 = 5 + 2(3) = 11\).
The terms are \(1, 1, 3, 5, 11\).
5 Make a conjecture for the general term of each sequence.
(a) \(1, 4, 9, 16, \ldots\) \(\qquad\) (b) \(2, 6, 12, 20, \ldots\)
(a) square numbers: \(u_n = n^2\)
(b) \(2 = 1 \times 2\), \(6 = 2 \times 3\), \(12 = 3 \times 4\), \(20 = 4 \times 5\), so \(u_n = n(n + 1)\).
6 Make a conjecture for the general term of \(\dfrac{1}{2},\; \dfrac{2}{4},\; \dfrac{3}{8},\; \dfrac{4}{16},\; \ldots\)
Numerators \(1, 2, 3, 4\) are \(n\); denominators \(2, 4, 8, 16\) are \(2^n\). So \(u_n = \dfrac{n}{2^n}\).
7 A sequence is defined by \(u_1 = 4\) and \(u_{n+1} = u_n + 3\).
(a) Make a conjecture for \(u_n\) in terms of \(n\).
(b) Hence find \(u_{50}\).
(a) \(4, 7, 10, 13, \ldots\): \(\;u_n = 4 + 3(n - 1) = 3n + 1\).
(b) \(u_{50} = 3(50) + 1 = 151\).
8 Sam owes $1000. At the end of each month, 1% interest is added to what he owes, and then he pays back $100. Let \(u_n\) be the amount he owes at the start of month \(n\), with \(u_1 = 1000\).
(a) Write down a recurrence relation for \(u_n\). \(\qquad\) (b) Find \(u_3\).
(a) \(u_1 = 1000\), \(\;u_{n+1} = 1.01\,u_n - 100\).
(b) \(u_2 = 1010 - 100 = 910\) and \(u_3 = 1.01(910) - 100 = 919.10 - 100 = 819.10\). He owes $819.10.
9 The sum of the first \(n\) terms of a series is \(S_n = n^2 + 4n\). Find \(u_1\) and a formula for \(u_n\).
\(u_1 = S_1 = 5\). For \(n \ge 2\):
\[ u_n = \left(n^2 + 4n\right) - \left((n-1)^2 + 4(n-1)\right) = n^2 + 4n - \left(n^2 + 2n - 3\right) = 2n + 3 . \]
This also gives \(u_1 = 5\) ✓.
10 The sum of the first \(n\) terms of a series is \(S_n = 2^{n+1} - 2\). Find \(u_n\), and hence \(u_5\).
\[ u_n = \left(2^{n+1} - 2\right) - \left(2^{n} - 2\right) = 2^{n+1} - 2^{n} = 2^n(2 - 1) = 2^n \]
(and \(u_1 = S_1 = 2 = 2^1\) ✓). So \(u_5 = 2^5 = 32\).