6. If a sequence of numbers $a_{1}, a_{2}, \cdots$ satisfies that for any positive integer $n$, 
$$
a_{1}+a_{2}+\cdots+a_{n}=n^{3}+3 n^{2},
$$

then $\frac{1}{a_{2}+2}+\frac{1}{a_{3}+2}+\cdots+\frac{1}{a_{2015}+2}=$ $\qquad$
$$
\text { II, } 6 . \frac{1007}{6048} \text {. }
$$

From the given information, we have
$$
\begin{array}{l}
a_{n}=n^{3}+3 n^{2}-(n-1)^{3}-3(n-1)^{2} \\
=3 n^{2}+3 n-2 .
\end{array}
$$

Thus, $\frac{1}{a_{n}+2}=\frac{1}{3 n^{2}+3 n}=\frac{1}{3 n(n+1)}$
$$
=\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+1}\right) \text {. }
$$

Therefore, the original expression is
$$
\begin{array}{l}
=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2015}-\frac{1}{2016}\right) \\
=\frac{1007}{6048} .
\end{array}
$$