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Algebra: Simultaneous Recurrence Relations

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Solve the simultaneous recurrence relations given below.

a(n) = 3a(n-1) + 2b(n-1)
b(n) = a(n-1) + 2b(n-1)

a(0) = 1
b(0) = 2

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Solution Summary

The expert solves a system of two recurrence sequences.

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We have the following system:

Firstly, we'll subtract the second equation from the first one, and get:
(1)
Now, if we put n-1 instead of n, we'll get:
(2)

Then, from the first equation, we'll subtract from both sides, and get:
- = 2( - ) (3).

And now we'll substitute (2) in (3) and get:
- = 4 , which is same as ...

Solution provided by:
Chris Busu, PhD (IP)
Education
  • BSc, University of Bucharest
  • MSc, Ovidius
  • MSc, Stony Brook
  • PhD (IP), Stony Brook
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