# 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!

• Using the formula for $F_n$ as a linear combination of geometric sequences, I think you should get an expression of your sum as a linear combination of geometric sums. Jul 11 '15 at 11:52
• What happens if you use Binet? Jul 11 '15 at 11:58
• Thanks for responding, @Joel. Yes, that approach would give an answer. But I was wondering if there was something clean - a nice factorised expression, like the first quoted result. Jul 11 '15 at 12:02

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.

• This is a nice answer, for sure ! Jul 11 '15 at 12:16
• Thanks for your efforts @Jack. I will accept this answer if no one is aware of an approach that might yield a factored expression, like the one quoted in the question. Jul 11 '15 at 12:16
• @JackD'Aurizio There is an error in the calculation. For demonstration take $n=1$. The summation is $-1$. The answer reads $(-7 - F_{3} - 4 F_{4} + 2 F_{2})/10 = (-7 - 2 -12 + 2)/10 = -17/10$. For the case of $n=2$ the summation is 1, whereas the solution is $-15/10$. Jul 11 '15 at 13:58
• @Leucippus: there must be some sign mistake, but the method is quite straightforward to follow, so feel free to fix it if you find the mistake before me. Jul 11 '15 at 14:11
• @Leucippus: it should be right now. Jul 11 '15 at 14:32

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}