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Old November 16th, 2008, 07:53 AM
cyb3r cyb3r is offline
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Default Recurrence relation

The problem:

Define the recurrence relation H(n) by the rules :
H(0) = 3 ,  H(1) = -3,  H(n) = 2H(n-1) - H(n-2) + 6 for n>=2. Calculate H(6). Prove by induction that H(n) is divisible by 3 for all n (n - positive integer).


I calculate H(6) using simple algebra and get H(6) = 57 (I will be grateful if you can check that, so I didn't miss anything). But I'm stuck on the proving by induction part. I need to show somehow that sum of the digits of the number H(n) are divisble by 3, but I don't know how to do it. I will appreciate any ideas or solutions, thank you
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