
2. For a given value $t$, we consider number sequences $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{n+1}=\frac{a_{n}+t}{a_{n}+1}$ for all $n \geqslant 1$.

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

(b) Suppose that $t=-3$. Investigate whether $a_{2020}=a_{1}$ for all starting values $a_{1}$ different from -1 and 1 .

## 2. Version for klas 6

(a) First, we determine for what starting values $a_{1}>0$ the inequalities $\frac{4}{3} \leqslant a_{2} \leqslant \frac{3}{2}$ hold. Then, we will prove that for those starting values, the inequalities $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$ are also valid for all $n \geqslant 2$.

First, we observe that $a_{2}=\frac{a_{1}+2}{a_{1}+1}$ and that the denominator, $a_{1}+1$, 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 therefore equivalent to the inequality

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

as we can multiply all parts in the inequality by the positive number $a_{1}+1$. Subtracting $a_{1}+2$ from all parts 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 therefore need to have $\frac{1}{3} a_{1} \leqslant \frac{2}{3}$ (i.e. $a_{1} \leqslant 2$ ), and $\frac{1}{2} \leqslant \frac{1}{2} a_{1}$ (i.e. $1 \leqslant a_{1}$ ). The starting 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 $\frac{4}{3} \leqslant a_{2} \leqslant \frac{3}{2}$. Looking at $a_{3}$, we see that $a_{3}=\frac{a_{2}+2}{a_{2}+1}$. That is the same expression as for $a_{2}$, only with $a_{1}$ replaced by $a_{2}$. Since $a_{2}$ also satisfies $1 \leqslant a_{2} \leqslant 2$, the same argument now shows that $\frac{4}{3} \leqslant a_{3} \leqslant \frac{3}{2}$.

We can repeat the same argument to show this for $a_{4}, a_{5}$, etcetera. Hence, we find that $\frac{4}{3} \leqslant a_{n} \leqslant \frac{3}{2}$ holds for all $n \geqslant 2$. The formal proof is done using induction: the induction basis $n=2$ has been shown above. For the induction step, see the solution of part (b) of the version for klas $5 \&$ klas 4 and below. The result is that all inequalities hold if and only if $1 \leqslant a_{1} \leqslant 2$.

(b) Let's start by computing the first few numbers of the sequence in terms of $a_{1}$. We see that

$a_{2}=\frac{a_{1}-3}{a_{1}+1} \quad$ and $\quad a_{3}=\frac{a_{2}-3}{a_{2}+1}=\frac{\frac{a_{1}-3}{a_{1}+1}-3}{\frac{a_{1}-3}{a_{1}+1}+1}=\frac{a_{1}-3-3\left(a_{1}+1\right)}{a_{1}-3+\left(a_{1}+1\right)}=\frac{-2 a_{1}-6}{2 a_{1}-2}=\frac{-a_{1}-3}{a_{1}-1}$.

Here, it is important that we do not divide by zero, that is, $a_{1} \neq-1$ and $a_{2} \neq-1$. The first inequality follows directly from the assumption. For the second inequality we consider when $a_{2}=-1$ holds. This is the case if and only if $a_{1}-3=-\left(a_{1}+1\right)$, if and only if $a_{1}=1$. Since we assumed that $a_{1} \neq 1$, we see that $a_{2} \neq-1$. The next number in the sequence is

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

Again, we are not dividing by zero since $a_{3}=-1$ only holds when $-a_{1}-3=-a_{1}+1$, which is never the case.

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

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

To conclude: indeed we have $a_{2020}=a_{1}$ for all starting values $a_{1}$ unequal to 1 and -1 .

