Sequence and Series

The Saturdays of a month fall on dates such as 3, 10, 17, 24 and 31. A child who saves Rs 10 more every week saves Rs 10, 20, 30, 40, … . A sheet of paper folded again and again has 1, 2, 4, 8, 16, … layers. Each of these is a list of numbers made by a rule, and the rule lets us find any number in the list without writing out all the numbers before it. Such a list is a sequence, and the sum of its numbers is a series. In this chapter you will learn what a sequence is, how to find its terms from a rule and its rule from its terms, and how to write long sums in a short form with the sigma sign \Sigma.

3.1 Sequence: meaning, terms and the general term

In the sequence 5, 10, 15, 20, \ldots we have t_1=5, t_2=10, t_3=15 and t_4=20. The number written small below t is the position of the term; the term itself is the value at that position. Keep the two apart: in this sequence the 4th term is 20.

Finite and infinite sequences

A sequence that stops is finite; one that goes on for ever is infinite. Three dots at the end, "…", show that the sequence continues.

Sequence	Type	Reason
3, 10, 17, 24, 31 (Saturdays of a month)	finite	it has exactly 5 terms
5, 10, 15, 20, … (multiples of 5)	infinite	there is no last multiple of 5
1, 2, 4, 8, …, 1024	finite	it stops at 1024
1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, …	infinite	the dots show it never ends

The general term

The general term (or nth term) t_n is a formula in n that gives the term in any position. For the multiples of 5, t_n=5n: the 1st term is 5\times1=5, the 7th term is 5\times7=35, and the 100th term is 500. Once we know t_n, we never need to count along the list.

A sequence is really a function whose domain is the set of natural numbers \{1,2,3,\ldots\}: it sends each position n to the term t_n. This is why the graph of a sequence is a set of separate dots, one above each whole number, and not a joined-up line.

Many sequences come from patterns of dots. Counting the dots in each picture of a growing pattern gives a sequence, and the shape of the pattern often shows the rule.

3.2 Finding terms and the rule of a sequence

From the rule to the terms

Any formula in n produces a sequence. The table shows some common shapes of general term and the sequences they make.

General term t_n	t_1	t_2	t_3	t_4	What happens
3n+1	4	7	10	13	goes up by 3 each time
n^2+1	2	5	10	17	gaps grow: 3, 5, 7
2^n	2	4	8	16	doubles each time
\frac{n}{n+1}	\frac{1}{2}	\frac{2}{3}	\frac{3}{4}	\frac{4}{5}	creeps up towards 1
(-1)^n\,2n	-2	4	-6	8	signs alternate

The factor (-1)^n is -1 when n is odd and +1 when n is even, so it makes the signs alternate. Use (-1)^{n+1} instead if the first term is to be positive.

Two kinds of rule

A sequence can be described in two ways.

Kind of rule	What it says	Example for 2, 5, 8, 11, \ldots
Term-to-term rule	how to get the next term from the one before	"start at 2 and add 3 each time": t_1=2, t_{n+1}=t_n+3
Position-to-term rule (general term)	how to get a term straight from its position	t_n=3n-1

The term-to-term rule is easy to see but slow to use: to reach the 100th term you must take 99 steps. The general term jumps there at once: t_{100}=299. That is why examinations ask for t_n.

Some famous sequences are known mainly by a term-to-term rule. In the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, \ldots every term after the second is the sum of the two terms before it: t_{n+2}=t_{n+1}+t_n. The numbers of spirals on a sunflower head or a pine cone are usually Fibonacci numbers.

From the terms to the rule: the method of differences

To find the rule, subtract each term from the next. The pattern of these differences tells you the kind of rule.

1. The first differences are all equal. If the gap is always d, the sequence is called an arithmetic sequence, and its general term has the form t_n=dn+c. To find c, compare with the multiples of d.

For 4, 7, 10, 13, \ldots the gap is 3, so compare with the multiples of 3:

n	1	2	3	4
3n	3	6	9	12
term	4	7	10	13

Each term is 1 more than 3n, so t_n=3n+1. The same idea gives the formula t_n=a+(n-1)d, where a is the first term: here 4+(n-1)3=3n+1.

2. The ratios are all equal. If each term is the previous one multiplied by the same number r, the sequence is a geometric sequence and t_n=a\times r^{n-1}. For 3, 6, 12, 24, \ldots we have a=3 and r=2, so t_n=3\times2^{n-1}.

3. The first differences change but the second differences are equal. Then the rule contains n^2. If the second difference is 2, compare the terms with n^2; if it is 2a, compare them with an^2.

Is a number a term, and which term is it?

Once t_n is known, set t_n equal to the number and solve for n. If n comes out as a natural number, the number is a term and n is its position. If n is a fraction or negative, the number is not a term.

Patterns that turn into sequences

3.3 Series and the sigma (Σ) notation

When the terms of a sequence are added, we get a series. The sequence 2, 4, 6, 8, 10 gives the series 2+4+6+8+10, whose value is 30. A series made from a finite sequence is a finite series; one made from an infinite sequence, such as 1+\frac{1}{2}+\frac{1}{4}+\cdots, is an infinite series.

An infinite series cannot be added term by term to the end, because there is no end. But its partial sums S_1, S_2, S_3, \ldots can still be worked out. For \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots they are

Each partial sum is closer to 1, but none of them goes past 1. Saying exactly what such a sum "approaches" is the idea of a limit, the subject of the next chapter.

The sigma notation

Writing 1+4+9+16+\cdots+400 in full is slow. Mathematicians use the Greek capital letter sigma, \Sigma (the Greek "S", for "sum"), to write such a sum in one short line:

Writing a series in sigma notation

Three steps: (1) find the general term t_k; (2) find the value of k for the first term (usually 1); (3) find the value of k for the last term by solving t_k= last term.

Rules for working with sigma

Because \Sigma only means "add", the ordinary rules of addition give three useful rules:

Rule	Example
\sum_{k=1}^{n}c=nc (a constant added n times)	\sum_{k=1}^{10}2=20
\sum_{k=1}^{n}c\,t_k=c\sum_{k=1}^{n}t_k	\sum_{k=1}^{10}3k=3\sum_{k=1}^{10}k
\sum_{k=1}^{n}(a_k+b_k)=\sum_{k=1}^{n}a_k+\sum_{k=1}^{n}b_k	\sum(3k+2)=\sum3k+\sum2

Two famous sums

When the young Carl Friedrich Gauss was asked to add the numbers from 1 to 100, he paired the first with the last, the second with the second-last, and so on: 1+100, 2+99, 3+98, … . Each pair makes 101, and there are 50 pairs, so the sum is 50\times101=5050. The picture below shows the same idea for any n.

From the sums back to the terms

If we know the sum S_n of the first n terms for every n, we can recover each term: the nth term is what the nth step adds,

Chapter summary

• A sequence is an ordered list made by a rule; its terms are t_1, t_2, \ldots, t_n. It is finite or infinite.
• The general term t_n gives the term in position n; a sequence is a function on the natural numbers.
• Equal first differences d: arithmetic, t_n=a+(n-1)d. Equal ratios r: geometric, t_n=ar^{n-1}. Equal second differences: the rule contains n^2.
• To test whether a number is a term, solve t_n= number; it is a term only if n is a natural number.
• A series is the sum of the terms; S_n=t_1+t_2+\cdots+t_n and t_n=S_n-S_{n-1}.
• \sum_{k=1}^{n}t_k means "add t_k for k=1, 2, \ldots, n"; the number of terms is upper - lower +1.
• \sum_{k=1}^{n}c=nc, \sum c\,t_k=c\sum t_k, \sum(a_k+b_k)=\sum a_k+\sum b_k.
• 1+2+\cdots+n=\frac{n(n+1)}{2} and 1+3+5+\cdots+(2n-1)=n^2.

Practice questions

Group A — Objective (1 mark each)

1. The 5th term of the sequence with t_n=2n+3 is [1]a10b13c11d8
2. The general term of 3, 6, 9, 12, \ldots is [1]an+3b3nc3n+3dn^3
3. The next term of 1, 4, 9, 16, \ldots is [1]a20b25c24d32
4. The sequence of even numbers 2, 4, 6, \ldots is [1]afinitebinfiniteca seriesdconstant
5. The common difference of 20, 16, 12, 8, \ldots is [1]a4b-4c8d12
6. The next term of 2, 6, 18, 54, \ldots is [1]a108b162c72d216
7. \sum_{k=1}^{4}k equals [1]a4b10c24d16
8. 2+4+6+8+10 in sigma notation is [1]a\sum_{k=1}^{5}2kb\sum_{k=2}^{10}kc\sum_{k=1}^{10}2kd\sum_{k=1}^{5}(k+2)
9. The number of terms in \sum_{r=3}^{10}t_r is [1]a7b8c10d13
10. If t_n=(-1)^n, then t_3 is [1]a1b-1c3d-3
11. The sum of the first 10 natural numbers is [1]a45b55c100d110
12. Which term of 5, 10, 15, \ldots is 60? [1]a10thb11thc12thd13th

Answers to Group A

1. (b) — t_5=2(5)+3=13.

2. (b) — the multiples of 3: t_n=3n.

3. (b) — square numbers; the next is 5^2=25.

4. (b) — there is no last even number.

5. (b) — each term is 4 less: 16-20=-4.

6. (b) — each term is 3 times the one before: 54\times3=162.

7. (b) — 1+2+3+4=10.

8. (a) — the kth term is 2k and there are 5 terms.

9. (b) — 10-3+1=8.

10. (b) — (-1)^3=-1.

11. (b) — \frac{10\times11}{2}=55.

12. (c) — 5n=60 gives n=12.

Group B — Short answer (2 marks each)

1. Write the first four terms of the sequence whose general term is t_n=n^2-1. [2]
2. Find the general term of 7, 11, 15, 19, \ldots. [2]
3. Find the 20th term of 2, 5, 8, 11, \ldots. [2]
4. Find the next two terms of 1, 3, 6, 10, 15, \ldots and state the rule. [2]
5. Is 75 a term of 4, 9, 14, 19, \ldots? Give a reason. [2]
6. Expand and evaluate \sum_{k=1}^{4}(3k-1). [2]
7. Write 1+8+27+64+125 in sigma notation. [2]
8. Find the value of \sum_{n=2}^{5}n^2. [2]
9. If t_n=\frac{n}{n+2}, find t_1, t_2, t_3 and t_{10}. [2]
10. Write \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} in sigma notation and find its value. [2]

Answers to Group B

1. t_1=0, t_2=3, t_3=8, t_4=15.

2. The difference is 4. Comparing with 4n (4, 8, 12, 16), each term is 3 more, so t_n=4n+3.

3. t_n=3n-1, so t_{20}=60-1=59.

4. The differences are 2, 3, 4, 5, so the next differences are 6 and 7: the next terms are 21 and 28. These are the triangular numbers, t_n=\frac{n(n+1)}{2}.

5. t_n=5n-1. Setting 5n-1=75 gives n=\frac{76}{5}=15.2, which is not a natural number. So 75 is not a term (the 15th term is 74 and the 16th is 79).

6. 2+5+8+11=26.

7. The terms are 1^3, 2^3, \ldots, 5^3, so the series is \sum_{k=1}^{5}k^3.

8. 4+9+16+25=54.

9. t_1=\frac{1}{3}, t_2=\frac{2}{4}=\frac{1}{2}, t_3=\frac{3}{5}, t_{10}=\frac{10}{12}=\frac{5}{6}.

10. The kth term is \frac{1}{2^k}, so the series is \sum_{k=1}^{4}\frac{1}{2^k}. Its value is \frac{8+4+2+1}{16}=\frac{15}{16}.

Group C — Long answer (4 marks each)

1. A row of n squares made of matchsticks uses 4, 7, 10, \ldots sticks for n=1, 2, 3, \ldots. (i) Find the general term. (ii) How many sticks are needed for 30 squares? (iii) How many squares can be made with 151 sticks? [4]
2. Find the general term of 2, 6, 12, 20, 30, \ldots by the method of differences, find its 15th term, and find \sum_{k=1}^{5}t_k. [4]
3. (i) Expand \sum_{k=1}^{6}(2k-1) and find its value. (ii) Show that the sum of the first n odd numbers is n^2 for n=1 to 6, and draw the square-and-L picture for n=4. [4]
4. The first row of a hall has 20 seats and each row has 3 more seats than the row in front. (i) Write the sequence and its general term. (ii) How many seats are in the 12th row? (iii) Which row has 62 seats? (iv) Write the number of seats in the first 5 rows in sigma notation and find it. [4]
5. The general term of a sequence is t_n=2^n-1. (i) Write its first five terms. (ii) Write the series of these five terms in sigma notation and find S_5. (iii) Is 255 a term? If so, which one? [4]

Answers to Group C

1. (i) The difference is 3; comparing with 3n, each term is 1 more, so t_n=3n+1. (ii) t_{30}=91 sticks. (iii) 3n+1=151 gives 3n=150, so n=50 squares.

2. First differences 4, 6, 8, 10; second differences 2, so compare with n^2: the leftovers 2-1=1, 6-4=2, 12-9=3, 20-16=4, 30-25=5 are n. So t_n=n^2+n. Then t_{15}=225+15=240, and \sum_{k=1}^{5}t_k=2+6+12+20+30=70.

3. (i) 1+3+5+7+9+11=36. (ii) Partial sums: 1=1^2, 1+3=4=2^2, 4+5=9=3^2, 9+7=16=4^2, 16+9=25=5^2, 25+11=36=6^2. For n=4 draw a 4\times4 square of dots split into L-shaped layers of 1, 3, 5 and 7 dots: each new odd number wraps one L around the square before it, making the next square.

4. (i) 20, 23, 26, 29, \ldots; the difference is 3, so t_n=3n+17. (ii) t_{12}=36+17=53 seats. (iii) 3n+17=62 gives 3n=45, so the 15th row. (iv) \sum_{k=1}^{5}(3k+17)=20+23+26+29+32=130 seats.

5. (i) 1, 3, 7, 15, 31. (ii) \sum_{n=1}^{5}(2^n-1)=1+3+7+15+31, so S_5=57. (iii) 2^n-1=255 gives 2^n=256=2^8, so n=8: 255 is the 8th term.