Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 8 - Section 8.2 - Solving Linear Recurrence Relations - Exercises - Page 525: 14

Answer

$α_n=2^n$

Work Step by Step

The characteristic equation of this recurrence relation is $$r^4 −5r^2 + 4 = 0$$ Because $r^4 −5r^2 + 4 = 0=(r^2 − 4)(r^2 − 1) = 0$. It follows that the solutions are $r_1 = 1, r_2 = -1, r_3=2, r_4=-2$ By Theorem 4 the solutions of this recurrence relation are of the form $α_n= α_1(-1)^n + α_21^n + α_3(-2)^n + α_42^n$ The simultaneous solutions are $α_1 =1, α_2 =1, α_3 = 0, α_4=1$ Therefore, $a_n = −1^n +1^n + 2^n = 2^n$
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