5. Let $\left\{x_{n}\right\}$ be a sequence of positive real numbers. If $x_{1}=\frac{3}{2}$ and $x_{n+1}^{2}-x_{n}^{2}=\frac{1}{(n+2)^{2}}-\frac{1}{n^{2}}$ for all positive integers $n$, find $x_{1}+x_{2}+\cdots+x_{2009}$.
(1 mark)設 $\left\{x_{n}\right\}$ 爲一個正實數數列。若 $x_{1}=\frac{3}{2}$, 且對所有正整數 $n$ 皆有 $x_{n+1}^{2}-x_{n}^{2}=\frac{1}{(n+2)^{2}}-\frac{1}{n^{2}}$, 求 $x_{1}+x_{2}+\cdots+x_{2009}$ 。
5. $\frac{4040099}{2010}$
5. For $n \geq 2$, we have
$$
\begin{aligned}
x_{n}^{2} & =x_{1}^{2}+\sum_{k=1}^{n-1}\left(x_{k+1}^{2}-x_{k}^{2}\right) \\
& =x_{1}^{2}+\sum_{k=1}^{n-1}\left[\frac{1}{(k+2)^{2}}-\frac{1}{k^{2}}\right] \\
& =\left(\frac{3}{2}\right)^{2}+\frac{1}{(n+1)^{2}}+\frac{1}{n^{2}}-\frac{1}{2^{2}}-\frac{1}{1^{2}} \\
& =\left(\frac{1}{n}-\frac{1}{n+1}+1\right)^{2}
\end{aligned}
$$

Since $x_{n}>0$, we have $x_{n}=\frac{1}{n}-\frac{1}{n+1}+1$ (note that this also holds for $n=1$ ). Hence
$$
\sum_{n=1}^{2009} x_{n}=\sum_{n=1}^{2009}\left(\frac{1}{n}-\frac{1}{n+1}+1\right)=1-\frac{1}{2010}+2009=\frac{4040099}{2010} .
$$

Remark. One could easily see a pattern and make a guess for the answer by computing $x_{1}$, $x_{1}+x_{2}, x_{1}+x_{2}+x_{3}$ and so on.