9. (10 points) As shown in the figure, the first polygon is "expanded" from an equilateral triangle, with the number of sides denoted as $a_{3}$, the second polygon is "expanded" from a square, with the number of sides denoted as $a_{4}, \cdots$, and so on. The number of sides of the polygon "expanded" from a regular $n$-sided polygon is denoted as $a_{n}(n \geqslant 3)$. If $\frac{1}{\mathrm{a}_{3}}, \frac{1}{\mathrm{a}_{4}}, \frac{1}{\mathrm{a}_{5}}, \cdots, \frac{1}{\mathrm{a}_{\mathrm{n}}}=\frac{2014}{6051}$, then $n=(\quad)$
(1)
(2)
(3)
(4)
A. 2014
B. 2015
C. 2016
D. 2017
【Solution】Solution: $a_{3}=3(2+2)=3 \times 4$,
$$
\begin{array}{l}
a_{4}=4(2+3)=4 \times 5, \\
a_{5}=5(2+4)=5 \times 6, \\
\cdots \\
a_{n}=n(n+1), \\
\therefore \frac{1}{3 \times 4}+\frac{1}{4 \times 5}+\frac{1}{5 \times 6}+\cdots+\frac{1}{n(n+1)}=\frac{2014}{6051}, \\
\therefore \frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{n}-\frac{1}{n+1}=\frac{2014}{6051}, \\
\therefore \frac{1}{3}-\frac{1}{n+1}=\frac{2014}{6051}, \\
\therefore n+1=2017, \\
\therefore n=2016 .
\end{array}
$$