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I am trying to solve following :

$$a(n)=2\cdot a(n-1)+5\cdot a(n-2)+6\cdot a(n-3)$$ with the initial conditions given by $a(0)=3,a(1)=2,a(2)=14$.

So first of all, I want to mark that there exists a theorem which states the following:

Let $P(x)=u_n\cdot x^n+u_{n-1}\cdot x^{n-1}+....+u_1\cdot x+u_0$ be a polynomial with real coefficients, and $u_n\neq 0$. If $P(x)$ has rational roots, they are of the form $\pm p/q$ where $p|u_0$ and $q|u_n$.

So first what I have tried is to construct characteristic equation or $$r^3-2\cdot r^2-5\cdot r-6=0$$ Now I have a problem to solve cubic equation also from the theorem I need a few help to understand it well.

Please help me. Thanks a lot.

  • 0
    It looks to me you can't avoid solving the cubic if you want an explicit solution. Note that the cubic has one real and two conjugate complex roots.2011-09-11
  • 0
    Where is this question from? I don't think there are rational roots to this equation.2011-09-11
  • 1
    The lazy way: [W|A](http://www.wolframalpha.com/input/?i=a%28n%29%3D2%E2%8B%85a%28n%E2%88%921%29%2B5%E2%8B%85a%28n%E2%88%922%29%2B6%E2%8B%85a%28n%E2%88%923%29%2Ca%280%29%3D3%2Ca%281%29%3D2%2Ca%282%29%3D14)2011-09-11
  • 0
    it is good Simon just wanted to see how it can be solved by some algebraic methods thanks2011-09-11
  • 3
    [This](http://math.stackexchange.com/questions/61725/61729#61729) should be of interest...2011-09-11
  • 0
    please post it as answer @ J. M2011-09-11
  • 1
    I am voting to close as a duplicate of the question J.M. linked, as this question is ***primarily about*** solving cubics, not solving recurrences.2014-06-27

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