10. (20 points) Given real numbers $a_{0}, a_{1}, \cdots, a_{2015}$, $b_{0}, b_{1}, \cdots, b_{2011}$ satisfy
$$
\begin{array}{l}
a_{n}=\frac{1}{65} \sqrt{2 n+2}+a_{n-1}, \\
b_{n}=\frac{1}{1009} \sqrt{2 n+2}-b_{n-1},
\end{array}
$$

where $n=1,2, \cdots, 2015$.
If $a_{0}=b_{2015}$, and $b_{0}=a_{2015}$, find the value of the following expression
$$
\sum_{k=1}^{2015}\left(a_{k} b_{k-1}-a_{k-1} b_{k}\right) .
$$
10. Notice that, for any $k=1,2 \cdots, 2015$, we have
$$
\begin{array}{l}
a_{k}-a_{k-1}=\frac{1}{65} \sqrt{2 k+2}, \\
b_{k}+b_{k-1}=\frac{1}{1009} \sqrt{2 k+2} .
\end{array}
$$

Multiplying the two equations, we get
$$
\begin{array}{l}
a_{k} b_{k}-a_{k-1} b_{k-1}+a_{k} b_{k-1}-a_{k-1} b_{k} \\
=\frac{2 k+2}{65 \times 1009} .
\end{array}
$$

Summing up, we have
$$
\begin{array}{l}
a_{2015} b_{2015}-a_{0} b_{0}+\sum_{k=1}^{2015}\left(a_{k} b_{k-1}-a_{k-1} b_{k}\right) \\
=\frac{2015(4+4032)}{2 \times 65 \times 1009}=62 .
\end{array}
$$

Therefore, the required value is 62.