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EXAMPLE 1

EXAMPLE 1. Evaluate recursive rules. Write the first six terms of the sequence. a. a 0 = 1, a n = a n – 1 + 4. b. a 1 = 1, a n = 3 a n – 1. SOLUTION. a. a 0 = 1. b. a 1 = 1. a 1 = a 0 + 4 = 1 + 4 = 5. a 2 = 3 a 1 = 3(1) = 3. a 2 = a 1 + 4 = 5 + 4 = 9.

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EXAMPLE 1

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  1. EXAMPLE 1 Evaluate recursive rules Write the first six terms of the sequence. a. a0 = 1, an= an – 1 + 4 b. a1 = 1, an= 3an – 1 SOLUTION a. a0 = 1 b. a1 = 1 a1 = a0 + 4 = 1 + 4 = 5 a2 = 3a1 = 3(1) = 3 a2 = a1 + 4 = 5 + 4 = 9 a3 = 3a2 = 3(3) = 9 a3 = a2 + 4 = 9 + 4 = 13 a4 = 3a3 = 3(9) = 27 a5 = 3a4 = 3(27) = 81 a4 = a3 + 4 = 13 + 4 = 17 a5 = a4 + 4 = 17 + 4 = 21 a6 = 3a5 = 3(81) = 243

  2. ANSWER So, a recursive rule for the sequence isa1 = 3, an= an– 1 + 10. EXAMPLE 2 Write recursive rules Write the first six terms of the sequence. a. 3, 13, 23, 33, 43, . . . b. 16, 40, 100, 250, 625, . . . SOLUTION The sequence is arithmetic with first term a1 = 3 and common difference d = 13 – 3 = 10. an= an – 1 + d General recursive equation for an = an – 1 + 10 Substitute 10 for d.

  3. b. The sequence is geometric with first term a1 = 16 and common ratio r = = 2.5. an= ran– 1 40 16 ANSWER So, a recursive rule for the sequence is a1 = 16,an= 2.5an – 1. EXAMPLE 2 Write recursive rules General recursive equation for an = 2.5an – 1 Substitute 2.5 for r.

  4. – 1. a1 = 3, an= an – 1 7 – – – a2 = a1 7 = 3 7 = 4 ANSWER 3, –4, –11, –18, –25 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. SOLUTION a1 = 3 a3 = a2– 7 a3 = – 4 – 7 = – 11 a4 = a3– 7 = – 11 – 7 = – 18 a5 = a4– 7 = – 18 – 7 = – 25

  5. ANSWER 162, 81, 40.5, 20.25, 10.125 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. 2. a0 = 162, an= 0.5an – 1 SOLUTION a0 = 162 a1 = 0.5a0 = 0.5 (162) = 81 a2 = 0.5a1 = 0.5 (81) = 40.5 a3 = 0.5a2 = 0.5 (40.5) = 20.25 a4 = 0.5a3 = 0.5 (20.25) = 10.125

  6. ANSWER 1, 2, 4, 7, 11 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. 3. a0 = 1, an= an – 1 + n SOLUTION a0 = 1 a1 = a0+ 1 = 1 + 1 = 2 a2 = a1+ 1 = 2+ 2 = 4 a3 = a2+ 3 = 4 + 3 = 7 a4 = a3+ 4 = 7 + 4 = 11

  7. a2 = 2a1– 1 = (2 4) – 1 = 8 – 1 = 7 a3 = 2a2– 1 = (2 7) – 1 = 14 – 1 = 13 a4 = 2a3– 1 = (2 13) – 1 = 26 – 1 = 25 ANSWER 4, 7, 13 25, 49 a5 = 2a4– 1 = (2 25) – 1 = 49 for Examples 1 and 2 GUIDED PRACTICE Write the first five terms of the sequence. 4. a1 = 4, an= 2an – 1– 1 SOLUTION a1 = 4

  8. an= ran– 1 a2 r = = 7 a1 ANSWER So, a recursive rule for the sequence is a1 = 2, an= 7an – 1 for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 5. 2, 14, 98, 686, 4802, . . . SOLUTION The sequence is geometric with first term a1 = 2 and common ratio = 7 ·an – 1

  9. ANSWER So, a recursive rule for the sequence is a1 = 19, and an= an – 1 – 6. for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 6. 19, 13, 7, 1, – 5, . . . SOLUTION The sequence is arithmetic with first term a = 19 and common difference a2– a1= – 6. an= an – 1 + d = an – 1 + (– 6)

  10. ANSWER So, a recursive rule for the sequence is a1 = 11, and an= an – 1 + 11. for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 7. 11, 22, 33, 44, 55, . . . SOLUTION The sequence is arithmetic with first term a = 11 and common difference a2– a1= 11. an= an – 1 + d = an – 1 + 11

  11. 1 a2 108 = 3 = a1 324 a = 324 = an= ran– 1 ANSWER 1 So, a recursive rule for the sequence is an = an – 1 3 1 a = 324, and an= an – 1 3 for Examples 1 and 2 GUIDED PRACTICE Write a recursive rule for the sequence. 8. 324, 108, 36, 12, 4, . . . SOLUTION The sequence is geometric with first term and common ratio

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