Sequences, terms and notation
A list of numbers runs on, or a pattern grows stage by stage. How do you name “the twentieth one” without writing out the first nineteen, and what kind of thing is such a list?
Where you are 10 in Sequences and series
Why this exists
Walk along one side of a street where the houses carry the odd numbers: 1, 3, 5, 7, 9 and on. Ask for “the fifth house” and the answer is house number 9, not 5. Two numbers are in play, the place in the row and the number on the door, and talking about the row without mixing them up needs a name for each.
Often you want one far-off entry, the 40th house, without writing out everything before it. A sequence is the idea that makes that possible: an ordered list, treated as a function of position.
Definition
A sequence is an ordered list of numbers. Each number in the list is a term. Each term sits in a position, counted 1, 2, 3 and so on from the start, and the position is called the index or term number, usually written n.
The term in position n is written a_n, read “a sub n”. For the odd house numbers, a_1 = 1, a_2 = 3, a_3 = 5 and a_5 = 9. The small n says where; the value a_n says what is there. Any letter will do in place of a, such as b_n or t_n, as long as each sequence keeps its own.
A sequence is a function. Put in a position n and out comes exactly one value, a_n. Its domain, the set of allowed inputs, is the counting numbers \{1, 2, 3, \ldots\}, because positions are counted. In function notation, f(n) means the same as a_n: for the house numbers, f(5) = 9.
- An infinite sequence has a term for every counting number. Three dots at the end, as in 1, 3, 5, 7, \ldots, say the list never stops.
- A finite sequence stops. Its domain is \{1, 2, \ldots, N\} for some last position N, and its list has no dots at the end.
Where counting starts. Some sequences are numbered from 0 instead, so the first term is a_0 and the domain is \{0, 1, 2, \ldots\}. That suits a list whose first entry is a starting value, before any step has happened. Neither choice is wrong, but they label the same list differently, so a sequence should always say where n starts. In this lesson n starts at 1 unless it says otherwise.
A sequence can be given as a list, as a table of n and a_n, as a picture pattern whose stages are counted, or as a graph. Often it also comes with a rule: a formula for a_n in terms of n, such as a_n = 2n - 1.
Intuition
Think of a sequence as a row of numbered boxes. The number on the front of a box is its position; the number inside is the term. Asking for a_{20} means “open box 20”.
Figure 1 draws the odd house numbers as a graph. Across is the position n, up is the value a_n, and each term is the point (n, a_n). There is a dot above 1, above 2, above 3 and so on, and nothing in between, because there is no house in position 2\tfrac12.
Results
Term number and term value. In a_n = v, the index n is the position and v is the value there. “The 5th term” means a_5, whatever number that turns out to be.
A rule gives any term directly. If a_n is given by a formula, put the position in for n: for a_n = 2n - 1, the 40th term is a_{40} = 2(40) - 1 = 79.
The graph of a sequence is the set of points (n, a_n), one above each number in the domain and no others. It is separate dots, never a joined line, and its domain, read off the graph, is the set of positions under the dots.
Why it is true
Why may a list be called a function? A function pairs each allowed input with exactly one output. A list does exactly that: position 1 holds one number, position 2 holds one number, and so on. Writing the list 1, 3, 5, 7, \ldots and writing the pairs (1, 1), (2, 3), (3, 5), (4, 7), \ldots say the same thing.
Why separate dots? The inputs are positions, and positions are counting numbers. A point on the graph at n = 2.5 would claim a term between the second and the third, and no such term exists. Joining the dots with a line would add infinitely many such false points. The dots may line up, as they do in Figure 1, but the sequence is only the dots.
Why does order matter? The lists 2, 5, 9 and 9, 5, 2 hold the same numbers, but the first has a_1 = 2 and the second has a_1 = 9. They give different outputs for the same input, so they are different sequences. A value may also repeat: 1, 1, 2, 2, 3, 3, \ldots is a perfectly good sequence, because two positions may hold the same value.
Worked examples
Term numbers and term values on a street
The houses on one side of a street are numbered with the odd numbers in order: 1, 3, 5, 7, \ldots Call the nth house number a_n. (a) Find a_5. (b) Explain why a_n = 2n - 1, and use it to find the number of the 20th house. (c) Which house in the row is number 35?
Solution(a) Count five places along the list: 1, 3, 5, 7, 9. So a_5 = 9. The fifth house is number 9: position 5, value 9.
(b) The nth even number is 2n, since the evens are 2 \times 1, 2 \times 2, 2 \times 3 and so on, and the nth odd number is one less than it. So a_n = 2n - 1. It agrees with the list: 2(1) - 1 = 1 and 2(5) - 1 = 9. Then
In function notation, with f(n) = 2n - 1, the same fact reads f(20) = 39.
(c) This time the value is known and the position is not. Set the rule equal to the value and solve:
House number 35 is the 18th house: a_{18} = 35.
A finite sequence: the Tuesdays in a month
A month has 31 days, and its first Tuesday falls on the 6th. Let t_n be the date of its nth Tuesday. List the sequence, and give its number of terms, its domain and t_3.
SolutionTuesdays come every 7 days, so each date is 7 more than the one before: 6, 13, 20, 27. The next would be 27 + 7 = 34, and the month has no 34th day. The list stops after four terms, so this is a finite sequence with domain \{1, 2, 3, 4\}. Reading the third place, t_3 = 20.
A rule fits the list: each date is 7n less 1, so t_n = 7n - 1, and indeed 7(1) - 1 = 6 and 7(4) - 1 = 27. The rule would happily give 7(5) - 1 = 34, but that is not a date. The formula does not decide where a finite sequence ends; the situation does, as in a reasonable domain for a real situation.
A picture pattern: matchstick squares
Figure 2 shows rows of squares made of matchsticks. Pattern n has n squares in a row, and m_n is its number of matchsticks. (a) Make a table of m_n for n = 1 to 5. (b) Find m_{10}. (c) Does any pattern use exactly 100 matchsticks? Exactly 200?
(a) Counting Figure 2, one square takes 4 matchsticks, two take 7 and three take 10. Each new square needs a top, a bottom and a right side, 3 matchsticks, because its left side is the right side of the square before. So the count goes up by 3 each time:
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| m_n | 4 | 7 | 10 | 13 | 16 |
(b) Pattern n is the first square’s 4 matchsticks plus 3 for each of the other n - 1 squares:
so m_{10} = 3(10) + 1 = 31.
(c) For 100: 3n + 1 = 100 gives 3n = 99, so n = 33, and pattern 33 uses exactly 100. For 200: 3n + 1 = 200 gives 3n = 199, so n = \tfrac{199}{3} = 66\tfrac13. A position must be a counting number, and there is no pattern 66\tfrac13. So 200 is not a term: pattern 66 uses 199 matchsticks and pattern 67 uses 202.
A sequence that adds the same amount every time, like this one, is called arithmetic, and arithmetic sequences have a lesson of their own.
Reading a sequence from a dot graph
Figure 3 shows the whole of a finite sequence a_n. Find a_2 and a_5, find the term whose value is 6, and give the domain.
Each dot is a point (n, a_n): go across to the position, then read the height.
Above n = 2 the dot is at height 5, so a_2 = 5. Above n = 5 the dot is at height 4, so a_5 = 4. The fifth term is 4, not 5.
The dot at height 6 stands above n = 4, so the term with value 6 is a_4, the fourth term.
The dots stand above 1, 2, 3, 4 and 5 and nowhere else, so the domain is \{1, 2, 3, 4, 5\}. As a list the sequence is 2, 5, 3, 6, 4. No formula is needed: a sequence only needs one value in each position.
Where the counting starts
A sequence starts at n = 0 and has the rule b_n = 3n + 5. (a) Write its first four terms. (b) Which term is the fourth one? (c) Find a rule a_n, starting at n = 1, that gives the same list.
Solution(a) The first position is 0, so b_0 = 3(0) + 5 = 5, then b_1 = 8, b_2 = 11 and b_3 = 14. The list is 5, 8, 11, 14, \ldots
(b) Counting from 0, the fourth term is b_3 = 14. The subscript is one less than the place in the list.
(c) Starting at 1 moves every label up by one, so a_n must equal b_{n-1}:
Check: 3(1) + 2 = 5 and 3(4) + 2 = 14. Same list and same fourth term, but labeled a_4 here and b_3 before. The formula alone does not show where n starts, which is why a sequence must say.
A rule that flips the sign
A sequence has the rule a_n = (-1)^n \cdot n. (a) Write its first six terms. (b) Find a_{25} and a_{40}. (c) Is 30 a term? Is -30?
Solution(a) The factor (-1)^n is -1 when n is odd and 1 when n is even, so it sets the sign, and the factor n sets the size:
| n | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| a_n | -1 | 2 | -3 | 4 | -5 | 6 |
No fixed amount is added: the gaps are 3, -5, 7, -9. No fixed amount multiplies either: the ratios are -2, -\tfrac32, -\tfrac43. Read as a function of n, though, the rule is plain.
(b) 25 is odd, so a_{25} = -25. 40 is even, so a_{40} = 40.
(c) The size of a_n is n, so a value of 30 or -30 could only sit in position 30. That position is even, so a_{30} = 30: 30 is a term and -30 is not. Same size, different answer, because the position decides the sign.
Boundaries
Positions are counting numbers. A term number is 1, 2, 3, \ldots, or 0, 1, 2, \ldots when the sequence says so. There is no a_{2.5}, no a_{-1}, and no a_0 in a sequence that starts at 1. When solving for a position gives a fraction or a negative, as with 200 matchsticks, the value is not a term.
Values are not restricted. A term can be negative, a fraction, or equal to another term. Only the positions have to be counting numbers.
A rule can run past the end. A formula for a finite sequence still produces numbers after the last position, like 34 for the Tuesdays. Those are not terms; the domain stops where the setting stops.
A few terms do not decide the rest. “Continue the pattern” assumes one particular rule, and a careful question names it. Closed form versus recursive form shows rules that agree at the start and then split.
Where this goes next
Most sequences worth studying come from a rule, and two kinds come first. Adding the same amount each time, like the matchsticks or the house numbers, makes an arithmetic sequence, whose rule is a straight-line formula in n: see arithmetic sequences and the explicit formula. Multiplying by the same amount each time, like the doubling list in Exercise 6, makes a geometric sequence, with n in the exponent: see geometric sequences and the explicit formula.
A rule can also build each term from the one before (“add 7 to the last Tuesday”) together with a starting value. That is a recursive formula, and generating a sequence from a rule puts the two kinds of rule side by side. Later, recurrence relations build each term from several earlier ones.
Exercises
Reading terms
- The sequence 4, 9, 14, 19, 24, \ldots goes up by 5 each time. Find a_1, a_2 and a_6, and say which term has the value 24.
- A sequence is given by this table. Find a_3 and a_4. For which n is a_n = 18? What is the 5th term?
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| a_n | 30 | 27 | 24 | 21 | 18 |
- Write the first five terms of the sequence a_n = 5n - 2.
- A sequence is written f(n) = 10 - 3n for n = 1, 2, 3, \ldots Find f(1), f(4) and f(10).
- Write the first four terms of a_n = n^2 - 4, and find a_{10}.
- Write the first five terms of a_n = 3 \cdot 2^{n-1}, and find a_6.
Patterns, lists and graphs
- A row of triangles is built from matchsticks, each new triangle sharing one side with the triangle before it. Pattern n has n triangles and uses s_n matchsticks, and pattern 1 uses 3. Find s_n for n = 1 to 4, find s_{10}, and find which pattern uses 41 matchsticks.
- On one side of a street the houses are numbered 1, 3, 5, \ldots, 49. How many houses are there, and what is the number of the 12th house?
- A month has 30 days and its first Tuesday falls on the 2nd. List the dates of its Tuesdays, give the number of terms and the domain, and find a rule for t_n.
- Figure 4 shows the whole of a finite sequence. Find a_2 and a_5, find which term has the value 9, give the domain, and find the one term whose value equals its own term number.
Notation and rules
- A sequence starts at n = 0 with b_n = 2n + 7. Write its first four terms. Which term is the fourth one, and what place in the list does b_4 hold?
- Write a rule for 6, 11, 16, 21, \ldots with n starting at 1, and another with n starting at 0.
- Is 60 a term of the sequence a_n = n^2 - 4? Is 40?
- Is 75 a term of the sequence a_n = 4n + 3? Is 100?
Reasoning
- Are 2, 5, 9 and 9, 5, 2 the same sequence? Explain using the idea of a function.
- Explain why the graph of the matchstick sequence in Example 3 is a row of dots and not a line.
- Explain why the dates of the Tuesdays in any month form a finite sequence, and why it always has four or five terms.
- A sequence has a_5 = 5. Must it be the sequence a_n = n? Explain with an example.
Answers
- a_1 = 4, a_2 = 9, a_6 = 29, and 24 is a_5. The sixth term is 5 more than the fifth, 24 + 5 = 29, and counting places shows 24 is fifth.
- a_3 = 24 and a_4 = 21. a_n = 18 when n = 5, so the 5th term is 18, not 5.
- 3, 8, 13, 18, 23. For example a_1 = 5(1) - 2 = 3 and a_5 = 5(5) - 2 = 23.
- f(1) = 7, f(4) = -2, f(10) = -20. Each is 10 minus three times the position: 10 - 3(4) = -2.
- -3, 0, 5, 12, and a_{10} = 96. Square the position and take away four: 10^2 - 4 = 96.
- 3, 6, 12, 24, 48, and a_6 = 96. For example a_1 = 3 \cdot 2^0 = 3 and a_6 = 3 \cdot 2^5 = 96. Each term is double the one before.
- 3, 5, 7, 9; s_{10} = 21; pattern 20. Each new triangle shares a side, so it needs only 2 new matchsticks: s_n = 3 + 2(n - 1) = 2n + 1. Then s_{10} = 2(10) + 1 = 21, and 2n + 1 = 41 gives n = 20.
- 25 houses, and the 12th is number 23. With a_n = 2n - 1, solving 2n - 1 = 49 gives n = 25, so 49 is the 25th term; and a_{12} = 2(12) - 1 = 23.
- 2, 9, 16, 23, 30: five terms, domain \{1, 2, 3, 4, 5\}, and t_n = 7n - 5. The next date would be 37, which is past the end of the month. Check: 7(1) - 5 = 2 and 7(5) - 5 = 30.
- a_2 = 8 and a_5 = 6; the value 9 is a_4; the domain is \{1, 2, 3, 4, 5, 6\}; and a_3 = 3. As a list the sequence is 5, 8, 3, 9, 6, 2, and only the third dot has height equal to its position.
- 7, 9, 11, 13. The fourth term is b_3 = 13, since counting starts at b_0 = 7. And b_4 = 15 is the fifth term.
- a_n = 5n + 1 starting at 1, and b_n = 5n + 6 starting at 0. Check: a_1 = 6, a_4 = 21, b_0 = 6, b_3 = 21. Both give the same list with different labels.
- Yes for 60: n^2 - 4 = 60 gives n^2 = 64, and the only position is n = 8 (-8 is not a position), so 60 = a_8. No for 40: n^2 = 44 has no whole-number solution, and indeed a_6 = 32 and a_7 = 45 jump past it.
- Yes for 75: 4n + 3 = 75 gives n = 18, so 75 = a_{18}. No for 100: 4n + 3 = 100 gives n = \tfrac{97}{4} = 24\tfrac14, which is not a counting number.
- No. A sequence is a function of position, and these two give different outputs at position 1: one has a_1 = 2, the other a_1 = 9. The same numbers in a different order make a different sequence.
- Positions are counting numbers. There is a pattern 2 and a pattern 3, but no pattern 2\tfrac12, which would be two and a half squares. A line through the dots would claim a value, 8.5 matchsticks, at n = 2.5, for a term that does not exist.
- The dates can go no higher than the last day of the month, so the list must stop. Any 28 days in a row are exactly four weeks and hold exactly four Tuesdays. A month has 28 to 31 days, so beyond its first 28 days there are at most 3 more, and 3 days in a row hold at most one Tuesday. So there are 4 or 5 terms.
- No. For example a_n = 2n - 5 has a_5 = 2(5) - 5 = 5, but a_1 = 2(1) - 5 = -3, not 1.