Example 3 Given $a_{n}=(2+\sqrt{7})^{2 n+1}, b_{n}$ is the fractional part of $a_{n}$. Then when $n \in \mathbf{Z}_{+}$, the value of $a_{n} b_{n}$ ( ). ${ }^{[1]}$
(A) must be an irrational number
(B) must be an even number
(C) must be an odd number
(D) can be either an irrational number or a rational number
$(2008$, National High School Mathematics League Jiangxi Province Preliminary Contest)
Notice that,
$$
\begin{aligned}
(\sqrt{7} & +2)^{2 n+1}-(\sqrt{7}-2)^{2 n+1} \\
& =2\left(\mathrm{C}_{2 n+1}^{0} 2^{2 n+1}+\mathrm{C}_{2 n+1}^{2} 2^{2 n-1} \times 7+\cdots+\right. \\
& \left.\mathrm{C}_{2 n+1}^{2 n} 2 \times 7^{n}\right) \in \mathbf{Z}_{+} .
\end{aligned}
$$

While $0<(\sqrt{7}-2)^{2 n+1}<1$, hence $(\sqrt{7}-2)^{2 n+1}$
is the fractional part of $a_{n}$, i.e.,
$$
b_{n}=(\sqrt{7}-2)^{2 n+1} \text {. }
$$

Then $a_{n} b_{n}=(\sqrt{7}+2)^{2 n+1}(\sqrt{7}-2)^{2 n+1}=3^{2 n+1}$ (an odd number). Therefore, the answer is C.