Problem 7. Let $(a_n)^{\infty}_{n=0}$ be the sequence defined recursively by $a_{n+3} = 6a_{n+2} - 11a_{n+1} + 6a_n$ for all $n \ge 0$, and such that $a_1 = 1$, $a_2 = 2$, and $a_3 = 1$. Find a closed form (i.e. non-recursive) expression for $a_n$.
I think it should be rewritten as $x^3 = 6x^2-11x+6$, but I am not sure how to apply $a_1$, $a_2$, and $a_3$. And so, I'm not sure how to find the closed form expression.
Any help?
$\textbf{Problem 7}$
Let $a_n:=x^n$
$$x^{n+3}-6x^{n+2}+11x^{n+1}-6x^{n}=0,$$
divide by $x^{n}$ to get
$$x^3-6x^2+11x-6=0,$$
doing the appropriate algebraic divisions yield
$$(x-3)(x-1)(x-2)=0 \Rightarrow x=\{3,1,2\}.$$
Then our general solution is of the form
$$a_n=A3^n+B+C2^n,$$
where $A,B,C$ are constants that we will find using the initial values.
For $a_1=1$ we have
$$1=3A+B+2C,$$
for $a_2=2$ we have
$$2=9A+B+4C,$$
and for $a_3=3$ we have
$$3=27A+B+8C.$$
Using your favourite method of solving linear systems, you should get (check!)
$$A=-\frac{1}{6}, B=-\frac{1}{2}, C=1.$$
Thus, the solution is
$$a_n=-\frac{1}{6}(3^n)-\frac{1}{2}+2^n.$$