Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.7 - Page 315: 7

Answer

ek = = (11(4k) – 5)/3
1519005884

Work Step by Step

ek = 4ek-1 + 5, for all integers k ≥ 1, e0 = 2 e0 = 2 e1 = 4(2) + 5 = 13 e2 = 4(4*2 + 5) + 5 = 57 e3 = 4(4*4*2 + 5*4 + 5) + 5 = 233 e4 = 4(4*4*4*2 + 5*4*4 + 5*4) + 5 = 2*44 + 5*43 + 5*42 + 5*41 + 5*40 = 2*4k + 5*4k-1 + 5*4k-2 + 5*4k-3 + 5*4k-4 = 2*4k + 5(4k-1 + 4k-2 + 4k-3 + 4k-4) = 2*4k + 5(∑_(i=0)^(k-1)▒4^i ) ek = 2*4k + 5(∑_(i=0)^(k-1)▒4^i ) = 2*4k + 5(4k-1+1 – 1)/(4-1) = 2*4k + 5(4k – 1)/3 = 2*4k + 5(4k)/3 – 5/3 = 6*4k/3 + 5(4k)/3 – 5/3 = (6(4k) + 5(4k) – 5)/3 = (11(4k) – 5)/3
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