10. Given the sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ satisfy: $a_{1}=-1, b_{1}=2, a_{n+1}=-b_{n}, b_{n+1}=2 a_{n}-3 b_{n}\left(n \in \mathbf{N}^{*}\right)$, then $b_{2015}+b_{2016}=$ $\qquad$ .
$$
\begin{array}{l}
a_{n}=-b_{n-1} \Rightarrow b_{n+1}=-2 b_{n-1}-3 b_{n} \Rightarrow b_{n}+b_{n+1}=-2\left(b_{n-1}+b_{n}\right), \\
\text { Also } b_{2}=2 a_{1}-3 b_{1}=-8 \Rightarrow b_{1}+b_{2}=-6 \text {, then } b_{n}+b_{n+1}=3 \cdot(-2)^{n} \text {. }
\end{array}
$$

So $b_{2015}+b_{2016}=3 \cdot(-2)^{2015}=-3 \cdot 2^{2015}$.