$10 \cdot 76$ If $T_{n}=1+2+3+\cdots+n$, and
$$
P_{n}=\frac{T_{2}}{T_{2}-1} \cdot \frac{T_{3}}{T_{3}-1} \cdot \frac{T_{4}}{T_{4}-1} \cdots \cdot \frac{T_{n}}{T_{n}-1},
$$

where $n=2,3,4, \cdots$, then the number closest to $P_{1991}$ is
(A) 2.0 .
(B) 2.3 .
(C) 2.6 .
(D) 2.9 .
(E) 3.2 .
(42nd American High School Mathematics Examination, 1991)
［Solution］From the given, we know
$$
\frac{T_{n}}{T_{n}-1}=\frac{\frac{n(n+1)}{2}}{\frac{n(n+1)}{2}-1}=\frac{n(n+1)}{n^{2}+n-2}=\frac{n(n+1)}{(n+2)(n-1)},
$$

then
$$
\begin{aligned}
P_{1991} & =\frac{2 \cdot 3}{4 \cdot 1} \cdot \frac{3 \cdot 4}{5 \cdot 2} \cdot \frac{4 \cdot 5}{6 \cdot 3} \cdots \frac{1990 \cdot 1991}{1992 \cdot 1989} \cdot \frac{1991 \cdot 1992}{1993 \cdot 1990} \\
& =\frac{3 \cdot 1991}{1993} \approx 2.997 .
\end{aligned}
$$

Therefore, the answer is $(D)$.