Expert: Sherman D. - 9/24/2005

Question
I think Advanced Math is the right place to put this... it's certainly advanced. I hope you can help me. :) I'm trying to solve the following recurrence relation: a(n) = 1/2a(n-1) + 1/2a(n-2) + n.
My approach has been to find a particular solution (the difficult part), and add it to the solution you find using the characteristic polynomial without the + n at the end. That didn't prove very hard at all. Since n is linear, I have been using a + bn to try to find a particular solution.  However, when I solve for b, I get b = b + 1, which isn't right at all.  I'm stuck.  Any suggestions you might have would be greatly appreciated.

Thank you,
Shawna

Answer
a(n) = (1/2)a(n - 1) + (1/2)a(n - 2) + n
a(n) = (a/2)((n - 1) + (n - 2)) + n
a(n) = (a/2)(n - 1 + n - 2) + n
a(n) = (a/2)(2n - 3) + n
a(n) = (a(2n - 3) + 2n)/2
a(n) = (2an - 3a + 2n)/2
a(n) = (2an + 2n - 3a)/2
a(n) = (2n(a + 1) - 3a)/2
a(n) = n(a + 1) - ((3a)/2)

maybe these sites can help.

www.ltcconline.net/greenl/Courses/103B/seqSeries/ARITSEQ.HTM

www.mathguide.com/lessons/SequenceGeometric.html