Expert: Paul Klarreich - 6/27/2006

Question
I'm working on problem here at work that I have boiled down to the following series...

F(1) = (1.03)^1(1.08)^0
F(2) = (1.03)^1(1.08)^1 + (1.03)^2(1.08)^0
F(3) = (1.03)^1(1.08)^2 + (1.03)^2(1.08)^1 + (1.03)^3(1.08)^0
F(4) = (1.03)^1(1.08)^3 + (1.03)^2(1.08)^2 + (1.03)^3(1.08)^1 + (1.03)^4(1.08)^0

I'm trying to solve for F(n), but I can't remember enough of my math courses in school to solve it any further than this.


Answer
Hi, Uriah,

I'm sorry -- this does not look familiar; the general expression has the form:

      k=n
F(n) = sum A^k B^(n-k)
      k=1

where, it seems, the zero term is missing.  This looks like one of those convolution-type expressions, but I don't know how to apply that here.