2. For a given value of $t$, we consider sequences of numbers $a_{1}, a_{2}, a_{3}, \ldots$ that satisfy the formula $a_{n+1}=\frac{a_{n}+t}{a_{n}+1}$ for all $n \geqslant 1$.

(a) Suppose that $t=2$. Determine all initial values $a_{1}>0$ such that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$ for all $n \geqslant 2$.

(b) Suppose that $t=-3$. Investigate whether for all initial values $a_{1}$ not equal to -1 and 1, it holds that $a_{2020}=a_{1}$.
## 2. Version for Grade 6

(a) First, we examine for which initial values $a_{1}>0$ it holds that $\frac{4}{3} \leqslant a_{2} \leqslant \frac{3}{2}$. Then we will prove that for all these initial values, the inequalities $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$ also hold for all other $n \geqslant 2$.

Note that $a_{2}=\frac{a_{1}+2}{a_{1}+1}$ and that the denominator here, $a_{1}+1$, is positive since $a_{1}>0$. The inequality

$$
\frac{4}{3} \leqslant a_{2}=\frac{a_{1}+2}{a_{1}+1} \leqslant \frac{3}{2}
$$

is thus equivalent to the inequality

$$
\frac{4}{3}\left(a_{1}+1\right) \leqslant a_{1}+2 \leqslant \frac{3}{2}\left(a_{1}+1\right),
$$

since you can multiply all terms in the inequality by the positive number $a_{1}+1$. If we then subtract $a_{1}+2$ from all terms of the inequality, we see that this is equivalent to

$$
\frac{1}{3} a_{1}-\frac{2}{3} \leqslant 0 \leqslant \frac{1}{2} a_{1}-\frac{1}{2} .
$$

We want $\frac{1}{3} a_{1} \leqslant \frac{2}{3}$, or $a_{1} \leqslant 2$, and $\frac{1}{2} \leqslant \frac{1}{2} a_{1}$, or $1 \leqslant a_{1}$. The initial value $a_{1}$ must therefore satisfy $1 \leqslant a_{1} \leqslant 2$.

Now suppose that $1 \leqslant a_{1} \leqslant 2$, so that $a_{2}$ satisfies $\frac{4}{3} \leqslant a_{2} \leqslant \frac{3}{2}$. If we now look at $a_{3}$, we see that $a_{3}=\frac{a_{2}+2}{a_{2}+1}$. This is exactly the same expression as that for $a_{2}$, only with $a_{2}$ instead of $a_{1}$. Since $a_{2}$ also satisfies $1 \leqslant a_{2} \leqslant 2$, it follows from the same argument that $\frac{4}{3} \leqslant a_{3} \leqslant \frac{3}{2}$.

We can repeat this argument to prove this for $a_{4}, a_{5}$, and so on. We thus find that for all $n \geqslant 2$ it holds that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$. The formal proof, of course, goes by complete induction: the base case $n=2$ we have seen above. For the proof of the induction step, we refer you to the solution of part (b) of the version for Grade 5 & Grade 4 and lower. The conclusion is that the required inequalities hold precisely when $1 \leqslant a_{1} \leqslant 2$.

(b) First, we calculate the first few values of the sequence in terms of $a_{1}$. We see that $a_{2}=\frac{a_{1}-3}{a_{1}+1} \quad$ and $\quad a_{3}=\frac{a_{2}-3}{a_{2}+1}=\frac{\frac{a_{1}-3}{a_{1}+1}-3}{\frac{a_{1}-3}{a_{1}+1}+1}=\frac{a_{1}-3-3\left(a_{1}+1\right)}{a_{1}-3+\left(a_{1}+1\right)}=\frac{-2 a_{1}-6}{2 a_{1}-2}=\frac{-a_{1}-3}{a_{1}-1}$.

It is important here that we do not divide by zero, i.e., that $a_{1} \neq-1$ and $a_{2} \neq-1$. The first inequality follows directly from the assumption, and for the second inequality, we examine when $a_{2}=-1$ holds. This holds precisely when $a_{1}-3=-\left(a_{1}+1\right)$, or when $a_{1}=1$. Since we have assumed that $a_{1} \neq 1$, we thus see that $a_{2} \neq-1$. The next term then becomes

$$
a_{4}=\frac{a_{3}-3}{a_{3}+1}=\frac{\frac{-a_{1}-3}{a_{1}-1}-3}{\frac{-a_{1}-3}{a_{1}-1}+1}=\frac{-a_{1}-3-3\left(a_{1}-1\right)}{-a_{1}-3+\left(a_{1}-1\right)}=\frac{-4 a_{1}}{-4}=a_{1} .
$$

Here, we also do not divide by zero, since $a_{3}=-1$ only holds if $-a_{1}-3=-a_{1}+1$, and this never holds.

We now see that $a_{4}=a_{1}$. Since $a_{n+1}$ depends only on $a_{n}$, we see that $a_{5}=a_{2}, a_{6}=a_{3}$, $a_{7}=a_{4}$, and so on. In other words: the sequence is periodic with period 3, and we see that

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

In conclusion: it indeed holds that $a_{2020}=a_{1}$ for all initial values $a_{1}$ not equal to 1 and -1.