10. Let the sequence $\left\{a_{n}\right\}$ satisfy $a_{1}=\sqrt{3}, a_{n+1}=\left[a_{n}\right]+\frac{1}{\left\{a_{n}\right\}}$, find the general term formula for $a_{n}$.
10. Analysis: It is easy to know that $a_{1}=\sqrt{3}, a_{2}=[\sqrt{3}]+\frac{1}{\{\sqrt{3}\}}=1+\frac{1}{\sqrt{3}-1}=\frac{3+\sqrt{3}}{2}$, so $a_{2} \in(2,3)$. Therefore, $a_{3}=2+\frac{2}{\sqrt{3}-1}=3+\sqrt{3}$, so $a_{3} \in(4,5), a_{4}=4+\frac{1}{\sqrt{3}-1}=\frac{9+\sqrt{3}}{2}$, and $a_{4} \in(5,6)$. Hence
(1) When $n=1,2$, it is obviously true.
(2) Assume that when $n=2 k-1,2 k(k \in \mathbb{Z})$, it is true. Then $a_{2 k}=\frac{6 k-3+\sqrt{3}}{2} \in(3 k-1,3 k)$. Therefore, $a_{2 k+1}=3 k-1+\frac{1}{\frac{6 k-3+\sqrt{3}}{2}-3 k+1}=3 k-1+\frac{2}{\sqrt{3}-1}=3 k+\sqrt{3}$, which is true. Also, $a_{2 k+1} \in(3 k+1,3 k+2)$, so $a_{2 k+2}=3 k+1+\frac{1}{\sqrt{3}-1}=\frac{6 k+3+\sqrt{3}}{2}$, which is also true. Therefore, by induction, we have: $a_{n}=\left\{\begin{array}{ll}\frac{3 n-3+2 \sqrt{3}}{2}, & n \text { is odd } \\ \frac{3 n-3+\sqrt{3}}{2}, & n \text { is even }\end{array}\right.$