1. Let $[x]$ denote the greatest integer not exceeding $x$. Then
$$
\begin{array}{l}
\frac{2}{[\sqrt{1 \times 2}] \times[\sqrt{2 \times 3}] \times[\sqrt{3 \times 4}]}+ \\
\frac{2}{[\sqrt{2 \times 3}] \times[\sqrt{3 \times 4}] \times[\sqrt{4 \times 5}]}+\cdots+ \\
\frac{2}{[\sqrt{98 \times 99}] \times[\sqrt{99 \times 100}] \times[\sqrt{100 \times 101}]} \\
\end{array}
$$

the value is
$$
\text { II. } 1 . \frac{4949}{9900} \text {. }
$$

Let $n$ be a positive integer, then
$$
n<\sqrt{n(n+1)}<n+1 \text {. }
$$

Thus, $[\sqrt{n(n+1)}]=n$.
Therefore, the original expression is
$$
\begin{array}{l}
=\frac{2}{1 \times 2 \times 3}+\frac{2}{2 \times 3 \times 4}+\cdots+\frac{2}{98 \times 99 \times 100} \\
=\frac{1}{1 \times 2}-\frac{1}{2 \times 3}+\frac{1}{2 \times 3}-\frac{1}{3 \times 4}+\cdots+ \\
\frac{1}{98 \times 99}-\frac{1}{99 \times 100} \\
=\frac{1}{1 \times 2}-\frac{1}{99 \times 100}=\frac{4949}{9900} .
\end{array}
$$