To determine The formula that expresses s n as a function of n by method of iteration, provided the recurrence relation and initial conditions. Explanation The given recurrence relation is, s n = 5 s n − 1 + 3 with the initial condition s 0 = 1 . Obtain the formula for s n by method of integration as follows. Substitute n = 1 in s n = 5 s n − 1 + 3 and obtain s 1 without simplifying the results to a numerical value as shown below. s 1 = 5 s 1 − 1 + 3 = 5 s 0 + 3 = 5 ( 1 ) + 3 Substitute n = 2 in s n = 5 s n − 1 + 3 and obtain s 2 follows. s 2 = 5 s 2 − 1 + 3 = 5 s 1 + 3 = 5 ( 5 ( 1 ) + 3 ) + 3 = 5 2 + 15 + 3 Now substitute n = 3 in s n = 5 s n − 1 + 3 and obtain s 3 as shown below. s 3 = 5 s 3 − 1 + 3 = 5 s 2 + 3 = 5 ( 5 2 + 15 + 3 ) + 3 = 5 3 + 75 + 15 + 3 Similarly substitute n = 4 in s n = 5 s n − 1 + 3 and obtain s 4 as shown below. s 4 = 5 s 4 − 1 + 3 = 5 s 3 + 3 = 5 ( 5 3 + 75 + 15 + 3 ) + 3 = 5 4 + 375 + 75 + 15 + 3 Using the above computations, the explicit formula for s n is computed as follows. s n = 5 n + 3 ( 5 ) n − 1 + 3 ( 5 ) n − 2 + … + 3 ( 5 ) + 3 = 5 n + 3 ( ( 5 ) n − 1 + ( 5 ) n − 2 + … + 5 + 1 ) = 5 n + 3 ( 1 + 5 + 5 2 + … + ( 5 ) n − 2 + ( 5 ) n − 1 )

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ISBN: 9780134689562

Author: Dossey, John A.

Publisher: Pearson,

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Chapter 9.2, Problem 17E

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The formula that expresses sn as a function of n by method of iteration, provided the recurrence relation and initial conditions.

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Chapter 9.2, Problem 16E

Chapter 9.2, Problem 18E

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Chapter 9 Solutions

Discrete Mathematics

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