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Situation: working on a homework problem for my discrete math class that I think is solved, but I am now wondering if my solution is right. This a is a recurrence relations problem with a quadratic equation to find the roots and then an extra term to deal with.

The question: Solve homogenous problem: $A_n = A_{n-1} + A_{n-2} + 2^n$ for $n \ge 2$.

My solution is: Roots $= \dfrac{1 \pm \sqrt{5}}{2}$

So I have $A_n = C_1\left( \dfrac{1 + \sqrt{5}}{2}\right)^n + C_2\left( \dfrac{1 - \sqrt{5}}{2}\right)^n $ where $C_1$ and $C_2$ are constants.

For the $2^n$, I have set the following: $A(2^n) - A(2^{n-1}) - A(2^{n-2}) = 2^n$, $A(1 -1/2 - 1/4) = 1$ then $A((4-2-1)/4) = 1$ or $A=4$.

For a particular solution, $A_n = 4(2)^n$.

Putting this all together, I have: $A_n = C_1\left( \dfrac{1 + \sqrt{5}}{2}\right)^n + C_2\left( \dfrac{1 - \sqrt{5}}{2}\right)^n + 4(2)^n$.

Am I on the right track? Any hints would be appreciated.

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    For the homogenous part (when you take away $2^n$), you have that $A_n$'s solution is constant (doesn't depend on $n$). – 2017-01-02
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    I reformatted this into MathJax. Is this what you want? – 2017-01-02
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    Yes this is it. Thanks for the formatting. – 2017-01-02

1 Answers 1

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Yeah! Seems quite good. Btw here's another nice way to eliminate the extra term and work out the recurrence:

Multiplying the giving recurrence by a factor of 2, seems like a hint to to me.

$$2A_n = 2A_{n-1} + 2A_{n-2} + 2^{n+1}$$

And also observe that

$$A_{n+1} = A_{n} + A_{n-1} + 2^{n+1}$$

From where we can easily eliminate the extra term and get $A_{n+1} - 3A_n + A_{n-1} + 2A_{n-2} = 0$ whose characteristic polynomial is $x^3-3x^2+x+2 = 0$ which in turn factorizes as $(x-2)(x-\varphi)(x+\varphi^{-1}) = 0$ where $\varphi = \frac{1+\sqrt{5}}{2}$ which is exactly what you got.

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    Thanks for the help! I like the alternative approach. – 2017-01-02
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    Pleasure was all mine! – 2017-01-02