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*}

Put $k =n-1$  and you get,
\begin{align}b_n &= 3^{n-1}b_1 + 3^{n-2} \cdot 5 + 3^{n-3} \cdot 5 + \dots +5 \\
 &= 2 \cdot 3^{n-1} + 5(1 + 3 + 3^2 + \dots + 3^{n-2}) \\
 &= 2\cdot 3^{n-1} + 5\left( \frac{3^{n-1}-1}{2}\right)\\
&= \frac{3^{n+1}-5}{2}
\end{align}
