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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,

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Presentation on theme: "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,"— Presentation transcript:

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3 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. 1234

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

5 As another example, consider the sequence given by a(n) = 3 n, or a n = 3 n. The notation a n means the same as a(n) but is used more commonly with sequences. Some function values (also called terms of the sequence) follow: a 1 = 3 1 = 3, a 2 = 3 2 = 9, a 3 = 3 3 = 27, a 4 = 3 4 = 81.

6 The first term of the sequence is a 1, the fifth term is a 5, and the nth term, or general term, is a n. This sequence can also be denoted in the following ways: 3, 9, 27,...; or3, 9, 27,..., 3 n,... The 3 n emphasizes that the nth term of this sequence is found by raising 3 to the nth power.

7 Example Solution Find the first four terms and the 15 th term of the sequence for which the general term is given by We have

8 Solution continued

9 Graphic Approach Mode y - editorTable Set Table Window Graph

10 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.

11 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 a n = n 3.

12 Solution continued b) 3,  9, 27,  81,... These are the powers of 3 with alternating signs, so the general term may be a n = (–1) (n + 1) (3 n ).

13 Sums and Series

14 Example Solution For the sequence 1, 5, 9, 13, 17, 21, 25, 29, find a) S 3 b) S 4 c) S 7 a) S 3 = 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) S 7 = 1 + 5 + 9 + 13 + 17 + 21 + 25 = 91 b) S 4 = 1 + 5 + 9 + 13 = 28

15 Sigma Notation When 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. This represents = (3·1 + 2) + (3·2 + 2) + (3·3 + 2).

16 Example Solution 2·1 2 + 2·2 2 + 2·3 2 = 2 + 8 + 18 = 28 Write out and evaluate each sum. a)b)c) a) Evaluate 2n 2 for all integers from 1 through 3. Then add.

17 Solution continued (2·3 – 3) + (2·4 – 3) + (2·5 – 3) 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

18 Graphing Calculator Use your calculator to find the first 8 terms for the sequence below. a) 2NDLIST5

19 Graphing Calculator Use your calculator to find the value of the sum. b) 2NDLIST5

20 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

21 Solution continued b) 4 + 8 + 16 + 32 + 64 Note that this is the sum of powers of 2, beginning with 2 2. One way to write this would be 2 n. The general term is 2 n + 1 beginning with n = 1. Sigma notation is


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