SOLUTION: The sequence u_n is defined recursively by the rules u_1 = 2, u_2 = 4, u_(n+2) = 4u_(n+1) + 5u_n for all n ∈ N. Prove the general term formula u_n = 5^n + (−1)^n.

Algebra ->  Test -> SOLUTION: The sequence u_n is defined recursively by the rules u_1 = 2, u_2 = 4, u_(n+2) = 4u_(n+1) + 5u_n for all n ∈ N. Prove the general term formula u_n = 5^n + (−1)^n.      Log On


   

Question 1033742: The sequence u_n is defined recursively by the rules
u_1 = 2, u_2 = 4, u_(n+2) = 4u_(n+1) + 5u_n for all n ∈ N.
Prove the general term formula u_n = 5^n + (−1)^n.
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Just to make a CORRECTION: I believe the formula you want to be proved is
u%5Bn%2B1%5D+=+5%5En%2B%28-1%29%5En
This is what we proceed to prove.
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Proof by induction:
If n = 3, the recurrence formula gives u%5B3%5D+=+4%2Au%5B2%5D%2B+5%2Au%5B1%5D+=+16%2B10+=+26.
Incidentally, the formula u%5B3%5D+=+5%5E2%2B%28-1%29%5E2+=+25%2B1+=+26 checks out.
Now assume that, up to a certain natural number k, u%5Bk%5D+=+5%5E%28k-1%29%2B%28-1%29%5E%28k-1%29 is true.
We show that u%5Bk%2B1%5D+=+5%5Ek%2B%28-1%29%5Ek.
Now u%5Bk%2B1%5D+=+4u%5Bk%5D+%2B5u%5Bk-1%5D
==>
=
=4%2A5%5E%28k-1%29-+4%2A%28-1%29%5E%28k-2%29+%2B+5%5E%28k-1%29%2B++5%2A%28-1%29%5E%28k-2%29
=5%5Ek+%2B+%28-1%29%5E%28k-2%29
=5%5Ek+%2B+%28-1%29%5Ek, since %28-1%29%5E2+=+1.
Therefore, u%5Bk%2B1%5D+=++5%5Ek+%2B+%28-1%29%5Ek, and the theorem is proved by induction.