Question: Done in mathematica # 4 What follows is called a second - order linear recurrence relation: a n = - 3 a n - 1

Done in mathematica
#4
What follows is called a second-order linear recurrence relation:
an=-3an-1-2an-2,n2
a0=0
a1=1
Write a recursive function in Mathematica that computes values for this recursion.
According to theorems about recursions, the solution to this recursion is an=2(-2)n-1+(-1)n,n0.
Test your recursion by comparing the value of a15 from your recursion and using the formula. Did they match?
 Done in mathematica #4 What follows is called a second-order linear

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