What is the $2006^{\text{th}}$ term in the sequence? In a sequence of integers, $t_1, t_2, \ldots; t_{n+3} = t_n+t_{n+1}-t_{n+2}$ for $n \geq 1$. If the first three terms, in order, are $1$, $3$, and $6$, what is the $2006^{\text{th}}$ term?
I thought this sequence would telescope and I would be able to find a closed-form expression for $t_n$ but that seems like it is not working. Also, there doesn't seem to be any pattern or period in the sequence.
 A: There are a few ways to do this, but here's a simple solution:
Let's compute the first few terms:
$1, 3, 6, -2, 11, -7, 16, -12, 21, ...$
You notice an alternating positive-negative pattern, which might suggest to just look at every other term. Starting from the first term, you get:
$1, 6, 11, 16, 21, ...$
And starting from the second:
$3, -2, -7, -12, ...$
There's a pretty clear pattern here: Both are arithmetic sequences. So we get the formulas:
$$t_{2n + 1} = 1 + 5n, \qquad n \geq 0$$
$$t_{2n} = 8 - 5n, \qquad n \geq 1$$
Once you have this formula, it is very easy to prove by induction. Now, using the second formula:
$$t_{2006} = 8 - 5(1003) = -5007$$
A: $t_{n+3}-t_{n+1} = -(t_{n+2}-t_n)=(-1)^n \cdot (t_{3}-t_1)=(-1)^n \cdot 5$
$\implies t_{n+3}=t_{n+1}+(-1)^n \cdot 5=t_{n-1}+((-1)^{n}+(-1)^{n-2}) \cdot 5= \cdots$
$\implies t_{2006}=t_{2004}+(-1)^{2003} \cdot 5=t_{2004}-5=\cdots =t_2-1002 \times 5=-5007$
A: Another more abstract, (and easier to prove) way is using linear algebra. Given $\begin{pmatrix}t_{n+2}\\t_{n+1}\\t_n\end{pmatrix}$, we can use the recursive formula in your question to see 
$$\begin{pmatrix}1&1&-1\\1&0&0\\0&1&0\end{pmatrix}\begin{pmatrix}t_{n+2}\\t_{n+1}\\t_n\end{pmatrix} =\begin{pmatrix}t_{n+3}\\t_{n+2}\\t_{n+1}\end{pmatrix}$$
So you just have to compute 
$$\begin{pmatrix}1&1&-1\\1&0&0\\0&1&0\end{pmatrix}^{2006}\begin{pmatrix}6\\3\\1\end{pmatrix}$$
The rest can be done with a calculator. To do it by hand you'd want to Jordan decompose the matrix. It doesn't hurt that bad. 
