Find the solution to each of these recurrence relations and initial conditions. Use an iterative approach such as that used in Example $10.$
- $a_{n} = 3a_{n−1}, a_{0} = 2$
- $a_{n} = a_{n−1} + 2, a_{0} = 3$
- $a_{n} = a_{n−1} + n, a_{0} = 1$
- $a_{n} = a_{n−1} + 2n + 3, a_{0} = 4$
- $a_{n} = 2a_{n−1} − 1, a_{0} = 1$
- $a_{n} = 3a_{n−1} + 1, a_{0} = 1$
- $a_{n} = na_{n−1}, a_{0} = 5$
- $a_{n} = 2na_{n−1}, a_{0} = 1$
