The 48th question, the integer sequence $\left\{a_{n}\right\}$ is defined as follows:
$$
a_{0}=3, a_{1}=4, a_{n+2}=a_{n+1} a_{n}+\left\lceil\sqrt{\left(a_{n+1}^{2}-1\right)\left(a_{n}^{2}-1\right)} \mid(n \geq 0)\right.
$$

Try to find the value of $\sum_{n=0}^{+\infty}\left(\frac{a_{n+3}}{a_{n+2}}-\frac{a_{n+2}}{a_{n}}+\frac{a_{n+1}}{a_{n+3}}-\frac{a_{n}}{a_{n+1}}\right)$. Here $[x]$ represents the smallest integer not less than $x$.
The 48th question,
Solution: It is easy to calculate that $\mathrm{a}_{2}=23$, and it is obvious that $\left\{\mathrm{a}_{\mathrm{n}}\right\}$ is a strictly increasing sequence of positive integers.
According to the condition, we have $a_{n+2}-a_{n+1} a_{n}=\left\lceil\sqrt{\left(a_{n+1}{ }^{2}-1\right)\left(a_{n}^{2}-1\right)}\right]$, which indicates that $\left(a_{n+2}-a_{n+1} a_{n}\right)^{2}$ is the smallest perfect square not less than $\left(a_{n+1}{ }^{2}-1\right)\left(a_{n}{ }^{2}-1\right)$.
We will prove by mathematical induction that:
$$
\left(a_{n+2}-a_{n+1} a_{n}\right)^{2}=\left(a_{n+1}^{2}-1\right)\left(a_{n}^{2}-1\right)+1
$$

When $\mathrm{n}=1$, the conclusion is obviously true.
Assume the conclusion holds for $\mathrm{n}$, i.e.,
$$
\begin{array}{l}
\left(a_{n+2}-a_{n+1} a_{n}\right)^{2}=\left(a_{n+1}^{2}-1\right)\left(a_{n}^{2}-1\right)+1 \\
\Rightarrow a_{n+2}^{2}+a_{n+1}{ }^{2}+a_{n}^{2}=2 a_{n+2} a_{n+1} a_{n}+2 \\
\Rightarrow a_{n}^{2}-\left(2 a_{n+2} a_{n+1}\right) a_{n}+\left(a_{n+2}{ }^{2}+a_{n+1}{ }^{2}-2\right)=0
\end{array}
$$

This indicates that $x=a_{n}$ is an integer root of the quadratic equation:
$$
x^{2}-\left(2 a_{n+2} a_{n+1}\right) x+\left(a_{n+2}^{2}+a_{n+1}^{2}-2\right)=0
$$

Thus, its discriminant
$$
\begin{array}{l}
\Delta=\left(2 a_{n+2} a_{n+1}\right)^{2}-4\left(a_{n+2}{ }^{2}+a_{n+1}{ }^{2}-2\right) \\
=4\left[\left(a_{n+2}{ }^{2}-1\right)\left(a_{n+1}{ }^{2}-1\right)+1\right]
\end{array}
$$

is a perfect square. Therefore, $\left(a_{n+2}^{2}-1\right)\left(a_{n+1}^{2}-1\right)+1$ is a perfect square and is clearly the smallest perfect square not less than $\left(a_{n+2} 2^{2}-\right.$ 1) $\left(\mathrm{a}_{\mathrm{n}+1}{ }^{2}-1\right)$. Hence,
$$
\left(a_{n+3}-a_{n+2} a_{n+1}\right)^{2}=\left(a_{n+2}{ }^{2}-1\right)\left(a_{n+1}^{2}-1\right)+1
$$

By mathematical induction, (1) holds.
From (1), for any non-negative integer $n$, we have:
$$
\begin{array}{l}
\left(a_{n+2}-a_{n+1} a_{n}\right)^{2}=\left(a_{n+1}^{2}-1\right)\left(a_{n}^{2}-1\right)+1 \\
\quad \Rightarrow a_{n+2}{ }^{2}+a_{n+1}{ }^{2}+a_{n}^{2}=2 a_{n+2} a_{n+1} a_{n}+2
\end{array}
$$

Thus, for any non-negative integer $n$, we have:
$$
\left\{\begin{array}{c}
a_{n+2}^{2}+a_{n+1}^{2}+a_{n}^{2}=2 a_{n+2} a_{n+1} a_{n}+2 \\
a_{n+3}{ }^{2}+a_{n+2}{ }^{2}+a_{n+1}{ }^{2}=2 a_{n+3} a_{n+2} a_{n+1}+2 \\
\Rightarrow a_{n+3} a_{n}^{2}-a_{n}^{2}=2 a_{n+3} a_{n+2} a_{n+1}-2 a_{n+2} a_{n+1} a_{n} \\
\Rightarrow a_{n+3}=2 a_{n+2} a_{n+1}-a_{n}
\end{array}\right.
$$

From (3), we know $\frac{a_{n+3}}{a_{n+2}}=2 a_{n+1}-\frac{a_{n}}{a_{n+3}}>2 a_{n+1}-1 \geq 2(n+1)-1>2 n$, so $\lim _{n \rightarrow+\infty} \frac{a_{n}}{a_{n+1}} \rightarrow 0$. Therefore, from (3), we have:
$$
\begin{array}{l}
\sum_{n=0}^{+\infty}\left(\frac{a_{n+3}}{a_{n+2}}-\frac{a_{n+2}}{a_{n}}+\frac{a_{n+1}}{a_{n+3}}-\frac{a_{n}}{a_{n+1}}\right) \\
=\frac{a_{3}}{a_{2}}-\frac{a_{2}}{a_{0}}+\frac{a_{1}}{a_{3}}-\frac{a_{0}}{a_{1}}+\sum_{n=1}^{+\infty}\left(\frac{a_{n+3}}{a_{n+2}}-\frac{a_{n+2}}{a_{n}}+\frac{a_{n+1}}{a_{n+3}}-\frac{a_{n}}{a_{n+1}}\right) \\
=\frac{a_{3}}{a_{2}}-\frac{a_{2}}{a_{0}}+\frac{a_{1}}{a_{3}}-\frac{a_{0}}{a_{1}}+\sum_{n=1}^{+\infty}\left(\frac{2 a_{n+2} a_{n+1}-a_{n}}{a_{n+2}}-\frac{2 a_{n+1} a_{n}-a_{n-1}}{a_{n}}+\frac{a_{n+1}}{a_{n+3}}-\frac{a_{n}}{a_{n+1}}\right) \\
=\frac{a_{3}}{a_{2}}-\frac{a_{2}}{a_{0}}+\frac{a_{1}}{a_{3}}-\frac{a_{0}}{a_{1}}+\sum_{n=1}^{+\infty}\left(-\frac{a_{n}}{a_{n}+2}+\frac{a_{n-1}}{a_{n}}+\frac{a_{n+1}}{a_{n+3}}-\frac{a_{n}}{a_{n+1}}\right) \\
=\frac{a_{3}}{a_{2}}-\frac{a_{2}}{a_{0}}+\frac{+\frac{a_{1}}{a_{3}}}{a_{3}}-\frac{a_{0}}{a_{1}}-\frac{a_{1}}{a_{3}}+\frac{a_{0}}{a_{1}} \\
=\frac{a_{3}}{a_{2}}-\frac{a_{2}}{a_{0}}=\frac{14}{69}
\end{array}
$$

In conclusion, the required sum $\sum_{n=0}^{+\infty}\left(\frac{a_{n+3}}{a_{n+2}}-\frac{a_{n+2}}{a_{n}}+\frac{a_{n+1}}{a_{n+3}}-\frac{a_{n}}{a_{n+1}}\right)=\frac{14}{69}$.