Sequences

Suppose that $5,000 is borrowed at 6%, compounded annually. The value of the loan at the start of the years 1, 2, 3, 4, and so on is

$5000, $5300, $5618, $5955.08, . . ., We can regard this as a function that pairs 1 with $5000, 2 with $5300, 3 with $5618, and so on. A sequence (or progression) is thus a function, where the domain is a set of consecutive positive integers beginning with 1, and the range varies from sequence to sequence.

1 2 3 4

If we continue computing the amounts in the account forever, we obtain an infinite sequence, with function values

$5000, $5300, $5618, $5955.08, $6312.38, . . .

The three dots at the end indicate that the sequence goes on without stopping. If we stop after a certain number of years, we obtain a finite sequence:

$5000, $5300, $5618, $5955.08

As another example, consider the sequence given by

a(n) = 3n, or an = 3n.

The notation an means the same as a(n) but is used more commonly with sequences. Some function values (also called terms of the sequence) follow:

a1 = 31 = 3,

a2 = 32 = 9,

a3 = 33 = 27,

a4 = 34 = 81.

The first term of the sequence is a1, the fifth term is a5, and the nth term, or general term, is an. This sequence can also be denoted in the following ways:

3, 9, 27, . . .;

or 3, 9, 27, . . ., 3n, . . .

The 3n emphasizes that the nth term of this sequence is found by raising 3 to the nth power.

Example

Solution

1 1(2 1) 2

12( 2)

a

Find the first four terms and the 15th term of the sequence for which the general term is given by 2

.( 2)n n

na

We have

2 2(2 2) 4

14( 2)

a

3 3(2 3) 6 3

8 4( 2)a

4 4(2 4) 8 1

16 2( 2)a

15 15(2 15) 30 15

32768 16384( 2)a

Solution continued

Graphic Approach

Mode y - editor Table Set

Table Window Graph

Finding the General Term

When only the first few terms of a sequence are known, it is impossible to be certain what the general term is. Still, a prediction can be made by looking for a pattern.

Example

Solution

For each sequence, predict the general term.

a) 1, 8, 27, 64, 125, . . .

b) 3, -9, 27, -81, . . .

a) 1, 8, 27, 64, 125, . . .

These are the cubes of consecutive positive integers, so the general term could be an = n3.

Solution continued

b) 3, -9, 27, -81, . . .

These are the powers of 3 with alternating signs, so the general term may be

an = (–1)(n + 1) (3n).

Sums and Series

Example

Solution

For the sequence 1, 5, 9, 13, 17, 21, 25, 29, find

a) S3 b) S4 c) S7

a) S3 = 1 + 5 + 9 = 15 This is the sum of the first 3 terms.

This is the sum of the first 4 terms.

This is the sum of the first 7 terms.

c) S7= 1 + 5 + 9 + 13 + 17 + 21 + 25 = 91

b) S4= 1 + 5 + 9 + 13 = 28

Sigma NotationWhen the general term of a sequence is known, the Greek letter Σ (capital sigma) can be used to write a series. For example, the sum of the first three terms of the sequence 5, 8, 11, 14, 17, . . . , 3k + 2, . . . can be named as follows, using sigma notation or summation notation:

This is read “the sum as k goes from 1 to 3 of (3k + 2).” The letter k is called the index of summation. The index of summation need not always start at 1.

3

1

(3 2)k

k

This represents

= (3·1 + 2) + (3·2 + 2) + (3·3 + 2) .

Example

Solution

2·12 + 2·22 + 2·32 = 2 + 8 + 18 = 28

Write out and evaluate each sum.

a) b) c)32

1

2n

n

5

3

(2 3)n

n

5

0

( 1) (2 3)k

k

k

a) 32

1

2n

n

Evaluate 2n2 for all integers from 1 through 3. Then add.

Solution continued

(2·3 – 3) + (2·4 – 3) + (2·5 – 3)

5

0

( 1) (2 3)k

k

k

5

3

(2 3)n

n

b)

c)

= (-1)0(2·0 + 3) + (-1)1(2·1 + 3) + (-1)2(2·2 + 3) + (-1)3(2·3 + 3) + (-1)4(2·4 + 3) + (-1)5(2·5 + 3)

= 3 + 5 + 7

= 15

= 3 – 5 + 7 – 9 + 11 – 13

= – 6

Graphing Calculator

Use your calculator to find the first 8 terms for the sequence below.

a) 22 2na n n

2ND LIST 5

Graphing Calculator

Use your calculator to find the value of the sum.

b)8

2

1

2 2n

n n

2ND LIST 5

Example

Solution

Write sigma notation for each sum.a) 1 + 5 + 9 + 13 b) 4 + 8 + 16 + 32 + 64

a) 1 + 5 + 9 + 13 Note that this is the sum of every other odd number or one

more than the multiples of four. One way to write this would be 4n + 1. The general term is (4n + 1), beginning with n = 0. Sigma notation is

4

1

(4 3)n

n

Solution continued

b) 4 + 8 + 16 + 32 + 64

Note that this is the sum of powers of 2, beginning with 22. One way to write this would be 2n. The general term is 2n + 1 beginning with n = 1. Sigma notation is

51

1

2n

n