Example 5 In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{n+1}=a_{n}+$ $\sqrt{a_{n}}+\frac{1}{4}$. Then $a_{99}=(\quad$.
(A) $2550 \frac{1}{4}$
(B) 2500
(C) $2450 \frac{1}{4}$
(D) 2401
Solution: From the recursive formula, we have $a_{n+1}=\left(\sqrt{a_{n}}+\frac{1}{2}\right)^{2}$, which means $\sqrt{a_{n+1}}=\sqrt{a_{n}}+\frac{1}{2}$.
Therefore, $\left\{\sqrt{a_{n}}\right\}$ is an arithmetic sequence. Thus, we have
$$
\sqrt{a_{n}}=\sqrt{a_{1}}+(n-1) \times \frac{1}{2}=\frac{n+1}{2} \text {. }
$$

Therefore, $a_{99}=\left(\frac{99+1}{2}\right)^{2}=2500$. Hence, the correct choice is (B).