<s>[INST] Solve the recurrence $b_1 = 2$ for $b_n = 3b_{n-1} + 5$ 
Solve the recurrence $b_1 = 2$ for $$b_n = 3b_{n-1} + 5$$

I've tried solving this problem using iteration, but the formula I get in the end is wrong. It is not a closed formula since there's still recurrence. I think my error is starts from the last line below. I'm not sure how to fix it. The formula I get in the end is:
$2(3^{n-1})+ 5((3^n - 1) / 3)$ 
which I got by setting $k = n -1$ and using $\sum_{k=0}^{n-1}3^k\ = (3^n-1)/2$
\begin{align*}
b_n&=3b_{n-1}+5\\
&=3(3b_{n-2}+5)+5\\
&=3^2b_{n-2}+3\cdot5+5\\
&=3^2(3b_{n-3}+5)+3\cdot5+5\\
&=3^3b_{n-3}+3^2\cdot5-3\cdot5+5\\
&\;\vdots\\
&=3^kb_{n-k}+3^{k-1}\cdot5+3^{k-2}\cdot5+\ldots+3\cdot5+5\\
\end{align*}
 [/INST] Let $a_{n} = b_n+c$ for some $c$ so that $a_{n+1}=3a_n$ (i.e. $a_n$ is geometric sequence). Then $$ a_{n+1}-c= 3a_n-3c+5\implies c = 5/2$$
and $$a_n= a_03^n$$ so $$ b_n = a_03^n-5/2$$
Since $b_1 = 2$ we get $3a_0=9/2$ so $a_0 =3/2$ so we finaly have  $$ b_n = {3^{n+1}-5\over 2}$$
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