Problem 4. Let $\left(a_{n}\right)_{n \geq 1}$ be a sequence of positive real numbers such that $a_{1}=1$ and

$$
\frac{1}{a_{1}+a_{2}}+\frac{1}{a_{2}+a_{3}}+\ldots+\frac{1}{a_{n-1}+a_{n}}=a_{n}-1, \text { for all } n \geq 2
$$

Determine the integer part of the number $A_{n}=a_{n} a_{n+1}+a_{n} a_{n+2}+a_{n+1} a_{n+2}$, where $n \in \mathbb{N}^{*}$.

Nelu Gerg, Slatina
Solution. For $n=2$, from the relation in the statement we obtain $a_{2}=\sqrt{2}$.

Writing the relation from the statement for $n \rightarrow n+1$, we get

$$
\frac{1}{a_{1}+a_{2}}+\frac{1}{a_{2}+a_{3}}+\ldots+\frac{1}{a_{n-1}+a_{n}}+\frac{1}{a_{n}+a_{n+1}}=a_{n+1}-1
$$

from which, using the statement again, it follows that for any $n \geq 2$ we have

$$
a_{n}-1+\frac{1}{a_{n}+a_{n+1}}=a_{n+1}-1 \Leftrightarrow\left(a_{n+1}-a_{n}\right)\left(a_{n+1}+a_{n}\right)=1 \Leftrightarrow a_{n+1}=\sqrt{a_{n}^{2}+1}
$$

By induction, it can be demonstrated that $a_{n}=\sqrt{n}$, for any $n \in \mathbb{N}^{*}$.

We have $A_{n}=\sqrt{n(n+1)}+\sqrt{n(n+2)}+\sqrt{(n+1)(n+2)}$. Since for any $n \in \mathbb{N}^{*}$ the following relations hold

$$
n+\frac{1}{3}<\sqrt{n(n+1)}<n+\frac{1}{2}, \quad n+\frac{2}{3}<\sqrt{n(n+2)}<n+1, \quad n+1<\sqrt{(n+1)(n+2)}<n+\frac{3}{2}
$$

by adding them, we obtain that $3 n+2<A_{n}<3 n+3$, from which $\left[A_{n}\right]=3 n+2$, for any $n \in \mathbb{N}^{*}$.