how do I prove this by induction? (recursion) The terms are given recursively:
$P_0=3$
$P_1=7$
and $P_n = 3P_{n-1}-2P_{n-2}$ for $n\ge2$
What should I assume and what step proves that $P_n=2^{n+2}-1$ is a closed form of the sequence.
Suppose $n_0=1$ and the base cases are $0$ and $1$.
I think this book has a mistake
problem from the book

Note that $2^{0+2}-1=2^{2}-1=4-1=3$ and $2^{1+2}-1=2^{3}-1=8-1=7$,
so that the base case holds. Suppose now that the statement holds
for $P_{n-1}$ and $P_{n-2}$ and note that
\begin{align*}
P_{n} & =3P_{n-1}-2P_{n-2}\\
 & =3\left(2^{n+1}-1\right)-2\left(2^{n}-1\right)\\
 & =3\cdot2^{n+1}-3-2^{n+1}+2\\
 & =2\cdot2^{n+1}-1\\
 & =2^{n+2}-1.
\end{align*}
