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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 Preview

    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 ...

    Solution Summary

    The expert solves a system of two recurrence sequences.