$\because U_{n+1}=3 U_n-2 U_{n-1}$
Step I Given, $U_1=3=2+1=2^1+1$, which is true for $n=1$, put $n=1$ in Eq. (i),
Then $U_{1+1}=3 U_1-2 U_{1-1}$
$$
\begin{aligned}
\Rightarrow \quad U_2 & =3 U_1-2 U_0 \\
& =3 \times 3-2 \times 2=5=2^2+1
\end{aligned}
$$
Which is true for $n=2$
$\therefore$ The result are true for $n=1$ and $n=2$
Step II Assume it is true for $n=k$, then it is also true for $n=k-1$
Then, $\quad U_k=2^k+1$
and $\quad U_{k-1}=2^{k-1}+1$
Step III On putting $n=k$ in Eq. (i), we get
$$
\begin{aligned}
U_{k+1} & =3 U_k-2 U_{k-1} \\
& =3\left(2^k+1\right)-2\left(2^{k-1}+1\right) \text { [From Eqs. (ii) and (iii)] } \\
& =3 \cdot 2^k+3-2 \cdot 2^{k-1}-2 \\
& =3 \cdot 2^k+3-2^k-2 \\
& =(3-1) 2^k+1 \\
& =2 \cdot 2^k+1=2^{k+1}+1
\end{aligned}
$$
This shows that the result is ture for $n=k+1$. Hence by the principle of mathematical induction the result is true for all $n \in N$.