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This is my first time working through this type of problem and I am looking to see if I have worked it out correctly, thanks!

Solve the reccurence relation $$a_n = 3a_{n-1}+4a_{n-2}, a_0=a_1 = 1$$

First we get the characteristic equation:

$$x^n = 3x^{n-1}+4x^{n-2}$$

Dividing by the smallest we get,

$$x^2=3x+4$$ $$x^2-3x-4 = 0 $$ $$x = 4,-1$$

Using the auxiliary equation we get the general solution:

$$a_n = A_1x^{n+1} + A_2x^{n+1}$$

Using conditions in the equation and our solution for $x$ gives,

$$a_0 = 1 = A_1(4) + A_2(-1) = 4A_1 - A_2$$ $$a_1 = 1 = A_1(4)^2 + A_2(-1)^2 = 16A_1 + A_2$$

Solving for $A_1, A_2$ I got $A_1 = \frac{1}{10}, A_2 = -\frac{3}{5}$

Therefore plugging into the general solution I got,

$$a_n =\left(\frac{1}{10}\right)(4)^{n+1}+\left(-\frac{3}{5}\right)(-1)^{n+1}$$

Did my workings come out correct?

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    $\begingroup$ Looks good but I would change one equation to $$a_n=A_1x_1^{n+1}+A_2x_2^{n+1}$$ because it's ambiguous as currently written. $\endgroup$ – user5713492 Mar 4 '18 at 18:12
  • $\begingroup$ good catch, thanks for that! $\endgroup$ – user123 Mar 4 '18 at 18:15
  • $\begingroup$ Sounds well to me! $\endgroup$ – Mostafa Ayaz Mar 4 '18 at 18:17
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    $\begingroup$ An easy way how to prove it is just to check if the values are right. So you have to take your result and calculate $a_0$ and $a_1$ and then try to put your solution into $a_{n-1}$ and $a_{n-2}$ to get your $a_n$. If you get those things right, then your solution has to be right. $\endgroup$ – HeatTheIce Mar 4 '18 at 21:06
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Yes it is correct and we can simplify a little as

$$a_n =\frac{2}{5}4^{n}+\frac{3}{5}(-1)^{n}$$

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    $\begingroup$ Can't this be better as a comment? $\endgroup$ – Jaideep Khare Mar 4 '18 at 18:20
  • $\begingroup$ The OP is asking for correctness of the solution, I’ve precisely answered to it and given an hint to simplify the final expression. What is the problem? $\endgroup$ – gimusi Mar 4 '18 at 18:24
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Yes it is correct. We can check it with the use of generating functions.
Transform $a_n = 3a_{n-1} + 4a_{n-2}$ to $$a_{n+2} = 3a_{n+1} + 4a_n \tag{1}$$.

Let $A(x) := \sum_{n \geq 0} a_n x^n$ then we get from (1)

$$ \begin{eqnarray*} \sum_{n \geq 0} a_{n+2} x^n &=& 3 \sum_{n \geq 0} a_{n+1} x^n + 4 \sum_{n \geq 0} a_n x^n \\ \Leftrightarrow \frac{A(x)}{x^2} - \frac{1}{x^2} - \frac{1}{x} &=& 3\Big(\frac{A(x)}{x} - \frac{1}{x}\Big) + 4 A(x) \\ \Leftrightarrow A(x) -1 -x &=& 3A(x)x-3x + 4A(x)x^2 \\ \Leftrightarrow A(x) &=& \frac{2x-1}{4x^2+3x-1} \end{eqnarray*} $$ With partial fractions we further obtain that $A(x) = \frac{3}{5(x+1)} + \frac{2}{5(1-4x)}$ and in series representation this is equivalent to $$ A(x) = \sum_{n \geq 0} x^n \Big(\frac{1}{5} 3(-1)^n + \frac{2}{5} 4^n \Big) $$ If we look now at the coefficient of $[x^n]$ in $A(x)$ we see that $$ a_n = \Big(\frac{1}{5} 3(-1)^n + \frac{2}{5} 4^n \Big) $$

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