SOLUTION: Define the a<sub>3</sub> and a<sub>4</sub> terms of the recurrence relation: a<sub>n+2</sub>= 2a<sub>n+1</sub> - 3a<sub>n</sub> where a<sub>0</sub> = a<sub>1</sub> = 3

Algebra.Com
Question 371452: Define the a3 and a4 terms of the recurrence relation:
an+2= 2an+1 - 3an where a0 = a1 = 3

Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!
an+2= 2an+1 - 3an where a0 = a1 = 3
Let n = 0
a0+2 = 2a0+1 - 3a0
   
a2 = 2a1 - 3a1
   
a2 = 2*3 - 3*3 = 6 - 9 = -3

Let n = 1

a1+2 = 2a1+1 - 3a1

a3 = 2a2 - 3a1

a3 = 2*-3 - 3*3 = -6 - 9 = -15

Let n = 2

a2+2 = 2a2+1 - 3a2

a4 = 2a3 - 3a2

a4 = 2*-15 - 3*-3 = -30 + 9 = -21

Keep going and you get:

a5 = 3
a6 = 69
a7 = 129
a8 = 51
a9 = -285
a10 = -723

Edwin
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