2. We consider sequences of numbers $a_{1}, a_{2}, a_{3}, \ldots$ that satisfy the formula $a_{n+1}=\frac{a_{n}+a_{1}}{a_{n}+1}$ for all $n \geqslant 1$.

(a) Suppose that $a_{1}=-3$. Calculate $a_{2020}$.

(b) Suppose that $a_{1}=2$. Prove that for all $n \geqslant 2$ it holds that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$.
## 2. Version for Grade 5 & Grade 4 and lower

(a) If $a_{1}=-3$, then $a_{2}=\frac{a_{1}+a_{1}}{a_{1}+1}=\frac{-6}{-2}=3$. Next, we calculate $a_{3}=\frac{a_{2}+a_{1}}{a_{2}+1}=\frac{0}{4}=0$. Then comes $a_{4}=\frac{a_{3}+a_{1}}{a_{3}+1}=\frac{-3}{1}=-3$. We see that $a_{4}=a_{1}$. Since $a_{n+1}$ depends only on $a_{n}$ and $a_{1}$, we see that $a_{5}=a_{2}=3, a_{6}=a_{3}=0, a_{7}=a_{4}=-3$, and so on. In other words: the sequence is periodic with period 3, and we see that

$$
a_{2020}=a_{2017}=a_{2014}=\cdots=a_{4}=a_{1}=-3 .
$$

(b) We will prove the statement by complete induction. We start with the base case $n=2$. We calculate $a_{2}=\frac{a_{1}+a_{1}}{a_{1}+1}=\frac{4}{3}$ and see that indeed $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$.

Now assume that for some $n=k \geqslant 2$ we have proven that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$. We will prove that these inequalities also hold for $n=k+1$.

First, we note that

$$
a_{k+1}=\frac{a_{k}+2}{a_{k}+1}=1+\frac{1}{a_{k}+1} .
$$

From the induction hypothesis $\left(\frac{4}{3} \leqslant a_{k} \leqslant \frac{3}{2}\right)$, it follows that $\frac{7}{3} \leqslant a_{k}+1 \leqslant \frac{5}{2}$ and thus that

$$
\frac{2}{5} \leqslant \frac{1}{a_{k}+1} \leqslant \frac{3}{7} .
$$

By adding 1 to each part of this inequality, we now find

$$
\frac{7}{5} \leqslant 1+\frac{1}{a_{k}+1}=a_{k+1} \leqslant \frac{10}{7} .
$$

Since $\frac{4}{3} \leqslant \frac{7}{5}$ and $\frac{10}{7} \leqslant \frac{3}{2}$, we see that $\frac{4}{3} \leqslant a_{k+1} \leqslant \frac{3}{2}$, as we wanted to prove.