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
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*}