# sol10 - AMS 301(Fall 2009 Homework Set 10 Solution Notes 6...

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AMS 301 (Fall 2009) Homework Set # 10 – Solution Notes # 6, 7.1: (a). a n = a n 1 + a n 2 . If the first digit is a 0, there are a n 1 ways to complete the sequence. If the first digit is a 1, then the second digit must be a 0, and then there are n - 2 digits left, so a n 2 ways to complete the sequence. a 1 = 2, a 2 = 3 are the initial condition. (b). a n = 2 a n 1 + 2 a n 2 . If the first digit is a 0, or a 2, there are a n 1 ways to complete the sequence. If the first digit is a 1, then the second digit must be a 0, or 2, and then there are n - 2 digits left, so a n 2 ways to complete the sequence. a 1 = 3, a 2 = 8 are the initial condition. # 24, 7.1: In this case once a digit 2 appears, all digits to its right must be 0 and 2’s (no 1’s allowed). Our recursion conditions on the first digit. If it is a 0 or 1 (two choices) then any sequence of length n - 1 that has no 1 to the right of a 2 can follow. However, if the first digit is a 2, the remaining n - 1 digits can each be a 0 or 2, so we have 2 n 1 such sequences. Therefore the recursion is:
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• Spring '08
• ARKIN
• Recurrence relation, Pallavolo Modena, Sisley Volley Treviso, Associazione Sportiva Volley Lube, Piemonte Volley, M. Roma Volley

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