Find the solution to the following linear, homogeneous recurrence with constant coefficients: an=8an−1−20an−2 for n≥2an=8an−1−20an−2

4 answers
Question:

Find the solution to the following linear, homogeneous recurrence with constant coefficients:
an=8an−1−20an−2 for n≥2an=8an−1−20an−2 for n≥2 with initial conditions a0=0,a1=−12a0=0,a1=−12. The solution is of the form:
an=(α+iβ)(r+is)n+(α−iβ)(r−is)nan=(α +iβ)(r+is)n+(α−iβ)(r−is)n
For suitable real constants α,β,r, sα,β,r, s.
Find these constants and enter their values:
α =
β =
r =
s =
sα =

Answers

step-by-step explanation:

first you will have to convert 3/4 into a decimal

which is .75

then you will have to divide 1.5 by .75

which is 2

no you have to divide

answer: the correct answer is -32

step-by-step explanation:

It is e) i am not sure if you have any questions or need any

oof

step-by-step explanation:

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