How can I prove that $a_n \le 2$ with induction for given progression?

Question

When given progression is: $$ {a_n}_{+1} = \frac{(a_n)^2}{4} + 1$$ $\forall n \in \mathbb{N}$ and $a_1 = 0$

Answer 1

hint

For the step $n+1$ (assuming that $a_n \le 2$)

$$a_{n+1}=\frac{(a_n)^2}{4}+1\le \frac{(2)^2}{4}+1=2$$

Now you just have to show that is valid for $n=1$.

Answer 2

Is $a_1 \le 2$? If $a_n \le 2$, is it true that $a_{n+1} \le 2$? What's $2^2$? What's $1+1$?

Answer 3

$$\begin{array} {rl} \text{Assumption} & a_n \in [0 ~ .. 2] \\ \text{Therefore} & a_n{}^2 \in [? ~ .. ?] \\ \text{Therefore} & \frac 14 a_n{}^2 + 1 \in [? ~ .. ?] \\ \text{Therefore} & a_{n + 1} \in [? ~ .. ?] \\ \end{array}$$

Figure out the $?$.