1. If $a, b$ are real numbers, and satisfy $(a-1)^{2}+$ $\sqrt{b-2}=0$, then
$$
\begin{array}{l}
\frac{1}{a b}+\frac{1}{(a+1)(b+1)}+\frac{1}{(a+2)(b+2)}+ \\
\cdots+\frac{1}{(a+2008)(b+2008)} \\
=(\quad) .
\end{array}
$$
(A) $\frac{2006}{2007}$
(B) $\frac{2007}{2008}$
(C) $\frac{2008}{2009}$
(D) $\frac{2009}{2010}$
$\begin{array}{l}\text { I.1.D. } \\ \text { Given }(a-1)^{2} \geqslant 0, \sqrt{b-2} \geqslant 0, \\ (a-1)^{2}+\sqrt{b-2}=0 \\ \Rightarrow a=1, b=2 . \\ \text { Therefore, the original expression }=\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{2009 \times 2010} \\ =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{2009}-\frac{1}{2010}\right) \\ =1-\frac{1}{2010}=\frac{2009}{2010} .\end{array}$