Alternating sum of product of Fibonacci numbers Suppose that $\{F_n\}$ is the sequence of Fibonacci numbers. There is a well-known result that
$$\sum_{i=1}^nF_i^2 F_{i+1}=\frac{1}{2}F_nF_{n+1}F_{n+2}.$$
This is easy to prove by induction. I was wondering if there known results for the corresponding alternating sum,
$$\sum_{i=1}^n (-1)^iF_i^2 F_{i+1}.$$
I've tried a few numerical hunts for a formula but couldn't find anything. Perhaps this problems has already been solved? Any hints or advice would be greatly appreciated! Ideal, would be an answer in terms of a factored product of Fibonacci and Lucas numbers, or something close!

An alternative process is the following.
By using 
\begin{align}
F_{n}^{2} \, F_{n+1} = \frac{1}{5} \left( F_{3n+1} - (-1)^{n} \, F_{n+2} \right)
\end{align}
then the desired series becomes
\begin{align}
S_{m} &= \sum_{n=1}^{m} (-1)^{n} \, F_{n}^{2} \, F_{n+1} \\
&= \frac{1}{5} \, \left[ \sum_{n=1}^{m} (-1)^{n} \, F_{3n+1} - \sum_{n=1}^{m} F_{n+2} \right] \\
&= \frac{1}{5} \left[ - \frac{1}{2} \left( F_{2} - (-1)^{m} F_{3m+2} \right) + F_{4} - F_{m+4} \right] \\
&= \frac{(-1)^{m} \, F_{3m+2} - 2 \, F_{m+4} + 5 }{10}.
\end{align}
