14.28 The sequence $\left\{a_{n}\right\}$ is defined as $a_{1}=1, \quad a_{n+1}=a_{n}+\frac{1}{a_{n}} \quad(n \geqslant 1)$. Find $\left[a_{100}\right]$.
(Japan Mathematical Olympiad, 1990)
[Solution] From $a_{n+1}^{2}=a_{n}^{2}+\frac{1}{a_{n}^{2}}+2$, we get

that is
$$
\begin{aligned}
& a_{n+1}^{2}-a_{1}^{2} \\
= & \sum_{i=1}^{n}\left(a_{i+1}^{2}-a_{i}^{2}\right) \\
= & 2 n+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}},
\end{aligned}
$$

that is
$$
a_{n+1}^{2}=2 n+1+\sum_{i=1}^{n} \frac{1}{a_{i}^{2}} \text {. }
$$

Thus, $\quad 199<a_{100}^{2}=199+\sum_{i=1}^{99} \frac{1}{a_{i}^{2}}$.
Since when $i \geqslant 2$, $a_{i}^{2} \geqslant 4$, hence
$$
199<a_{100}^{2} \leqslant 200+\frac{98}{4}<225 .
$$

Therefore, $14^{2}<a_{100}^{2}<15^{2}$.
So $\left[a_{100}\right]=14$.