How do I find an explicit formula, using backtracking, for $C_n =3C_{n-1} -2C_{n-2}$ with initial values $C_1=5$ and $C_2=3$? I have the recurrence $C_n =3C_{n-1} -2C_{n-2}$ with initial values $C_1=5$ and $C_2=3$, and want to find an explicit formula for $C_n$.
I made a table with the terms up to the $6$th term and noticed:
\begin{align}
C_3 &= C_2-4\\
C_4 &= C_3-8\\
C_5 &= C_4-16\\
C_6 &= C_5-32
\end{align}
But I still can't find the correct formula.

Those four equations you wrote mean: $$C_n = C_{n-1} -2^{n-1}.$$
Then you can use the recurrence directly: \begin{align} C_n &= C_{n-1} -2^{n-1}\\ &=C_{n-2} -2^{n-2} -2^{n-1}\\ &=C_{n-3} -2^{n-3} -2^{n-2} -2^{n-1}\\ &\vdots\\ &= C_1 -\sum_{k=1}^{n-1} 2^k\\ &= C_1 \bbox[yellow]{+1} -\sum_{k=\bbox[yellow]{0}}^{n-1} 2^k\\ &= C_1 +1 -(2^n -1)\\ &= C_1 +2 -2^n \qquad\qquad(C_1=5)\\ &=7 -2^n.\end{align} 
