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!
 A: If we use the explicit formula:
$$ F_n = \frac{1}{\sqrt{5}}\left(\sigma^n-\overline{\sigma}^n\right),\qquad \sigma+\overline{\sigma}=1,\quad \sigma\overline{\sigma}=-1,$$
then we have:
$$ \sum_{i=1}^{n} (-1)^i F_i^2 F_{i+1} =\\= \frac{1}{5\sqrt{5}}\sum_{i=1}^{n}(-1)^i\left[\left(\sigma^{3i+1}-\overline{\sigma}^{3i+1}\right)-2(-1)^i\left(\sigma^{i+1}-\overline{\sigma}^{i+1}\right)-(-1)^{i+1}\left(\sigma^{i-1}-\overline{\sigma}^{i-1}\right)\right]=\\=\frac{1}{5}\sum_{i=1}^{n}\left[(-1)^i F_{3i+1}-2F_{i+1}+F_{i-1}\right]=\frac{-2(F_{n+3}-2)+(F_{n+1}-1)}{5}+\frac{1}{5}\sum_{i=1}^{n}(-1)^i F_{3i+1}$$
and we just need to compute the last sum:
$$\begin{eqnarray*} \frac{1}{5\sqrt{5}}\sum_{i=1}^{n}(-1)^i\left(\sigma^{3i+1}-\overline{\sigma}^{3i+1}\right) &=&\frac{1}{5\sqrt5}\left[\frac{\sigma^2}{2}(-1+(-1)^n \sigma^{3n})-\frac{\overline{\sigma}^2}{2}(-1+(-1)^n \overline{\sigma}^{3n})\right]\\&=&\frac{1}{10}\left(-1+(-1)^n F_{3n+2}\right)\end{eqnarray*} $$
to have:

$$\sum_{i=1}^{n}(-1)^i F_i^2 F_{i+1} = \frac{5+(-1)^n F_{3n+2}-4 F_{n+3}+2 F_{n+1}}{10}.$$

As a curiosity, if we plug in $n=6$ we get $666$. $37$ is a divisor of $666$, but the first Fibonacci number that is divisible by $37$ is $F_{19}$, that is $F_{3n+1}$ in this case. So the chances that the alternating sum can be written as a nice product are very slim.
A: 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}
