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$$a_n=4a_{n-1}+5a_{n-2},\quad a_1=2,a_2=6$$

$$x^2-4x-5=0$$ $$x=-2+i,-2-i$$(complex roots)
as per the quadratic equation for the roots, $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ Then what is the formula ?

thank for confirming the roots just need one more clarification, after substituting these roots my a_n=C_1(-1)^n+C_2(5)^n =>a_1=C_1(-1)^1+C_2(5)^1 =>a_1=-C_1+5C_2 similarly for a_2=C_1(-1)^2+C_2(5)^2 =>a_2=C_1+25C_2 If this solution correct.

Monica
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  • What formula do you need? I mean, the reccurence formula is $a_n=4a_{n-1} + 5a_{n-2}$. – 5xum Feb 12 '14 at 07:38
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    The solution of your quadratic equation is wrong. $x=-1$ is an obvious root and $x=5$ is the second root. $b^2-4a c=16+20=36$. I suppose you missed a sign. – Claude Leibovici Feb 12 '14 at 07:43
  • Actually, even with complex roots, you just use the formula for the general solution you have been using. However, as Claude Leibovici mentions, your discriminant is 16+20, so your roots are real. – Eric Towers Feb 12 '14 at 07:56
  • http://en.wikipedia.org/wiki/Recurrence_relation#Solving – lab bhattacharjee Feb 12 '14 at 07:57

1 Answers1

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Beside the fact that the roots you gave in your post are wrong, the solution to your problem is fully described in the link lab bhattacharjee provided in his comment.

So, learn (if still required) the method and apply it. You should arrive to $$a(n)=\frac{2}{15} \left(2\ 5^n-5 (-1)^n\right)$$

Claude Leibovici
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