Given that `u_(n+1)=3u_n-2u_(n-1),` and `u_0=2 ,u_(1)=3`, then prove that `u_n=2^(n)+1` for all positive integer of `n`

Explanation

To prove that \( u_n = 2^n + 1 \) for all positive integers \( n \) given the recurrence relation \( u_{n+1} = 3u_n - 2u_{n-1} \) with initial conditions \( u_0 = 2 \) and \( u_1 = 3 \), we will use the principle of mathematical induction.

### Step 1: Base Case
We first verify the base cases for \( n = 0 \) and \( n = 1 \).

- For \( n = 0 \):
\[
u_0 = 2^0 + 1 = 1 + 1 = 2
...