Closed Form Solution for a Recurrence Relation We start out with: $a_n = 3a_{n-1} - 7, a_0 = 2$. Is the following valid?
$$a_{n-1} = 3(3a_{n-2}-7)-7 \\ a_{n-1} = 3^2a_{n-2}-(3\cdot-7) - 7 \\ \vdots \\ a_k = 3^ka_{n-k} -(3^{k-1}\cdot-7) \cdots -7\\ \text{Let $k = n$} \\ a_n = 3^ka_0 - (3^{n-1}\cdot-7)-\cdots-7$$
I don't know where to go from here. I would I make a closed form out of this?

As Greg Martin has already pointed out, your closed form cannot be right, because it doesn’t satisfy the recurrence. You could also see whether it generates the right values for $a_1,a_2$, and $a_3$, say, and find that it doesn’t.
There is a slightly better way to organize this sort of ‘unwinding’ of a simple recurrence, but make sure that you do the algebra correctly:
$$\begin{align*}
a_n&=3a_{n-1}-7\\
&=3(3a_{n-2}-7)-7\\
&=3^2a_{n-2}-3\cdot7-7\\
&=3^2(3a_{n-3}-7)-3\cdot7-7\\
&=3^3a_{n-3}-3^2\cdot7-3\cdot7-7\\
&\;\;\vdots\\
&=3^ka_{n-k}-7\sum_{i=0}^{k-1}3^i\\
&\;\;\vdots\\
&=3^na_0-7\sum_{i=0}^{n-1}3^i\\
&=2\cdot3^n-7\cdot\frac{3^n-1}{3-1}\\
&=2\cdot3^n-\frac{7\cdot3^n-7}2\\
&=\frac12(7-3\cdot3^n)\\
&=\frac12(7-3^{n+1})
\end{align*}$$
Note that it’s generally best not to do too much simplification at each stage: too much simplification tends to obscure the pattern. And when you’re done, you should always check to be sure that your closed form satisfies the recurrence:
$$3\left(\frac12(7-3^n)\right)-7=\frac32\cdot7-\frac12\cdot3^{n+1}-7=\frac12(3^{n+1}-7)\,.$$
