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
We have the following system:
Firstly, we'll subtract the second equation from the first one, and get:
Now, if we put n-1 instead of n, we'll get:
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 ...
The expert solves a system of two recurrence sequences.