<s>[INST] 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?
 [/INST] 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:
$$3\left(\frac12(7-3^n)\right)-7=\frac32\cdot7-\frac12\cdot3^{n+1}-7=\frac12(3^{n+1}-7)\,.$$
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