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What is the explicit formula of the sequence with $$ \begin{cases} a_0=1\\ a_1=2\\ a_n=3a_{n-1}+4a_{n-2} & \text{if }n\ge 2 \end{cases} .$$

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    Any thoughts? Do you know how to work with the characteristic polynomial of a linear recursion? – 2017-01-06
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    What technique(s) have you studied for solving such recurrences? The elementary method using the auxiliary equation $x^2-3x-4=0$ works very nicely here. – 2017-01-06
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    Have you tried computing some values of $a_n$ maybe you can see a pattern. – 2017-01-06
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    See https://math.stackexchange.com/questions/1540681/solving-the-recurrence-relation-a-k-3a-k-14a-k-2-with-a-0-1-a-1-2?rq=1 – 2017-09-28

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This is kind of like the Fibonacci sequence, but a little different. You can solve it quite easily by noticing that

$$a_{n+2}=3a_{n+1}+4a_n\\x^2=3x+4\implies\begin{cases}x_1=4\\x_2=-1\end{cases}$$

Thus,

$$a_n=y_1(x_1)^n+y_2(x_2)^n=y_1\times4^n+y_2\times(-1)^n$$

With $a_0=1$ and $a_1=2$, we can deduce that

$$a_n=\frac35\times4^n+\frac25\times(-1)^n$$

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    When $n=0$ this solution gives $a_0 = \frac15$. – 2017-01-07
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    Oops, thanks for the proof reading! Typos... – 2017-01-07
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Recursions of this sort can be soolved by the ansatz that $a_n = x^n$ and the realization that the recursion is linear, so that if you have two solutions, you can combine them to form a third.

This trick gives you a polynomial equation in $x$; in this case the equation is $$ x^2 = 3x + 4 $$ which has solutions $x=4$ and $x=-1$. So the recursion has a general solution of $$ a_n = \alpha\cdot 4^n + \beta\cdot(-1)^n$$

The initial conditions ($a_0=1,a_1=2$) let you find that $\alpha = \frac35,\beta = \frac25$ so the answer is

$$ a_n = \frac35\cdot 4^n +\frac25\cdot(-1)^n$$