6 Given the sequence $\left\{a_{n}\right\}$ with the general term formula $a_{n}=\frac{1}{(n+1) \sqrt{n}+n \sqrt{n+1}}(n \in$
$\left.\mathbf{N}^{*}\right)$, and its first $n$ terms sum as $S_{n}$, then in the sequence $S_{1}, S_{2}, \cdots, S_{2008}$, the number of rational terms is ).
(A) 43 terms
(B) 44 terms
(C) 45 terms
(D) 46 terms
6 Because
$$
\begin{aligned}
a_{k} & =\frac{1}{\sqrt{k(k+1)}(\sqrt{k+1}+\sqrt{k})} \\
& =\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}} \\
& =\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}(k=1,2, \cdots),
\end{aligned}
$$

So
$$
\begin{aligned}
S_{n} & =\sum_{k=1}^{n} a_{k} \\
& =\sum_{k=1}^{n}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right) \\
& =1-\frac{1}{\sqrt{n+1}}
\end{aligned}
$$

Also, since $\sqrt{2009}<45$, we have $n+1=2^{2}, 3^{2}, \cdots, 44^{2}$. Therefore, among $S_{1}, S_{2}, \cdots, S_{2008}$, there are 43 rational terms. Hence, the answer is A.