Let $F_n$ denote the $n^{th}$ Fibonacci number, given by $F_1=F_2=1$ and $F_n=F_{n-1}+F_{n-2}$. Suppose that $$\sum_{x=1}^{1000} F_xF_{x+2017}$$ can be expressed as $\frac{F_a+F_b-F_c-F_d}{e}$ where $e<10$. Compute $a+b+c+d+e$.
Is there a way that we can solve this using recursion. I know that the Fibonacci number sequence depends on recursion to operate. Help is much appreciated and needed.
Note: I'm looking for any possible solution.
This answer uses that $$F_nF_{m}=\frac 15(L_{m+n}-(-1)^nL_{m-n})\tag1$$ $$L_{n-1}+L_{n+1}=5F_n\tag2$$ $$F_{2n}=F_{n+1}^2-F_{n-1}^2=(F_{n+1}-F_{n-1})(F_{n+1}+F_{n-1})=F_n(F_{n+1}+F_{n-1})\tag3$$ $$\sum_{i=1}^{n}F_{2i-1}^2=\frac{1}{5}\left(3F_{2n}^2+2F_{2n-1}^2-4F_{2n-2}F_{2n}+2n-2\right)\tag4$$ $$F_{m+1}F_{n+1}-F_{m-1}F_{n-1}=F_{m+n}\tag5$$ $$3F_{n+2}-2F_n=F_{n+2}+2(F_{n+2}-F_{n})=F_{n+2}+2F_{n+1}=F_{n+3}+F_{n+1}\tag6$$
The proof for $(2)$ can be seen here.
The proof for $(3)$ can be seen here.
The proof for $(4)$ can be found in this answer.
The proofs for $(1)(5)$ are written at the end of the answer.
Using $(1)(2)(3)(4)(5)(6)$, we have $$\small\begin{align}&\sum_{x=1}^{1000}F_xF_{x+2017}\\\\&\stackrel{(1)} =\frac 15\left(\sum_{x=1}^{1000}L_{2x+2017}\right)-\frac{L_{2017}}{5}\sum_{x=1}^{1000}(-1)^x\\\\&=\frac 15\sum_{x=1}^{1000}L_{2x+2017}\\\\&=\frac 15\sum_{x=1}^{500}\left(L_{4x+2015}+L_{4x+2017}\right)\\\\&\stackrel{(2)} =\sum_{x=1}^{500}F_{4x+2016}\\\\&\stackrel{(3)} =\sum_{x=1}^{500}(F_{2x+1009}^2-F_{2x+1007}^2)\\\\&=\sum_{i=1}^{1005}F_{2i-1}^2-\sum_{i=1}^{505}F_{2i-1}^2-\sum_{i=1}^{1004}F_{2i-1}^2+\sum_{i=1}^{504}F_{2i-1}^2 \\\\&\stackrel{(4)} =\frac 15\left(3F_{2010}^2+2F_{2009}^2-4F_{2008}F_{2010}+2008\right)-\frac 15\left(3F_{1010}^2+2F_{1009}^2-4F_{1008}F_{1010}+1008\right)\\\\&\qquad -\frac 15\left(3F_{2008}^2+2F_{2007}^2-4F_{2006}F_{2008}+2006\right)+\frac 15\left(3F_{1008}^2+2F_{1007}^2-4F_{1006}F_{1008}+1006\right) \\\\&=\frac 15\left(3(F_{2010}^2-F_{2008}^2)-3(F_{1010}^2-F_{1008}^2)+2(F_{2009}^2-F_{2007}^2)-2(F_{1009}^2-F_{1007}^2)\right)\\\\&\qquad +\frac 15\left(-4(F_{2008}F_{2010}-F_{2006}F_{2008})+4(F_{1008}F_{1010}-F_{1006}F_{1008})\right) \\\\&\stackrel{(3)(5)} =\frac 15\left(3F_{4018}-3F_{2018}+2F_{4016}-2F_{2016}-4F_{4016}+4F_{2016}\right)\\\\&=\frac 15\left((3F_{4018}-2F_{4016})-(3F_{2018}-2F_{2016})\right)\\\\&\stackrel{(6)} =\frac 15\left(F_{4019}+F_{4017}-F_{2019}-F_{2017}\right)\end{align}$$
So, $a+b+c+d+e=4019+4017+2019+2017+5=12077$.
Finally, let us prove $$F_nF_{m}=\frac 15(L_{m+n}-(-1)^nL_{m-n})\tag1$$ $$F_{m+1}F_{n+1}-F_{m-1}F_{n-1}=F_{m+n}\tag5$$
Proof for $(1)$ :
Let us prove $(1)$ by induction on $n$.
It holds for $n=0$ since the both sides are equal to $0$.
It holds for $n=1$ since
$$F_1F_{m}=F_m=\frac 15(L_{m+1}+L_{m-1})=\frac 15(L_{m+1}-(-1)^1L_{m-1})$$ where we used $(2)$.
Suppose that it holds for $n-1,n$.
Then, $$\begin{align}F_{n+1}F_m&=(F_n+F_{n-1})F_m\\\\&=F_nF_m+F_{n-1}F_m\\\\&=\frac 15(L_{m+n}-(-1)^nL_{m-n})+\frac 15(L_{m+n-1}-(-1)^{n-1}L_{m-(n-1)})\\\\&=\frac 15(L_{m+n}+L_{m+n-1}-(-1)^{n+1}(-L_{m-n}+L_{m-n+1}))\\\\&=\frac 15(L_{m+n+1}-(-1)^{n+1}L_{m-(n+1)})\qquad\blacksquare\end{align}$$
Proof for $(5)$ :
Let us prove by induction on $n$ that $$F_{m+1}F_{n+1}-F_{m-1}F_{n-1}=F_{m+n}\tag5$$
It holds for $n=1$ since $F_{m+1}F_{1+1}-F_{m-1}F_{1-1}=F_{m+1}$.
It holds for $n=2$ since $$F_{m+1}F_{2+1}-F_{m-1}F_{2-1}=2F_{m+1}-F_{m-1}=F_{m+1}+F_{m}=F_{m+2}$$
Suppose that it holds for some $n-1,n$.
Then, $$\begin{align}F_{m+1}F_{n+2}-F_{m-1}F_{n}&=F_{m+1}(F_{n+1}+F_n)-F_{m-1}(F_{n-1}+F_{n-2})\\\\&=(F_{m+1}F_{n+1}-F_{m-1}F_{n-1})+(F_{m+1}F_n-F_{m-1}F_{n-2})\\\\&=F_{m+n}+F_{m+n-1}\\\\&=F_{m+n+1}\qquad\blacksquare\end{align}$$
$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}} $, with $\varphi = \frac{1+\sqrt{5}}{2}$, $\psi = \frac{1-\sqrt{5}}{2}$ (implying $\varphi=-\psi^{-1}$), so we get \begin{equation} \begin{split} F_x F_{x+2017} & = \frac{1}{5}\left(\varphi^{x}\varphi^{x+2017} - \varphi^{x} \psi^{x+2017} - \psi^{x} \varphi^{x+2017} + \psi^{x}\psi^{x+2017} \right)\\ & = \frac{1}{5}\left(\varphi^{2x+2017} - (-1)^x(\psi^{2017} + \varphi^{2017}) + \psi^{2x+2017} \right). \end{split} \end{equation}
Summing up $F_xF_{x+2017}$, the middle terms vanish due to alternating sings of constant expressions, leaving \begin{equation} \begin{split} \sum_{x=1}^{1000} F_xF_{x+2017} =\; & 5^{-1}\sum_{x=1}^{1000}(\varphi^{2x+2017}+\psi^{2x+2017}) \\ =\; & 5^{-1}\left(\varphi^{2017}\sum_{x=1}^{1000}\varphi^{2x}\right)+ 5^{-1}\left(\psi^{2017}\sum_{x=1}^{1000}\psi^{2x}\right). \end{split} \end{equation} With the geometric sum formula $\sum_{x=1}^{1000} \xi^{2x} = \frac{1-\xi^{2002}}{1-\xi^2} -1 $ this yields \begin{equation} \begin{split} & \;5^{-1}\left(\varphi^{2017}\left( \frac{1-\varphi^{2002}}{1-\varphi^2} -1 \right) + \psi^{2017}\left( \frac{1-\psi^{2002}}{1-\psi^2} -1 \right) \right)\\ = &\; \frac{ (1-\psi^2)(\varphi^{2019}-\varphi^{4019}) + (1-\varphi^2)(\psi^{2019}-\psi^{4019})}{5(1-\psi^2)(1-\varphi^2)}\\ = &\; \frac{\varphi^{2019}-\varphi^{4019} - \varphi^{2017}+\varphi^{4017} + \psi^{2019}-\psi^{4019} - \psi^{2017}+\psi^{4017}}{-5}. \end{split} \end{equation} Writing this using Lucas numbers ($L_n = \varphi^n + \psi^n$), we get $$\frac{1}{5}[L_{4019}-L_{4017} - (L_{2019}-L_{2017})].$$ $L$ and $F$ are related by $L_n = F_{n+1} + F_{n-1}$ yielding $L_{n+2} - L_{n} = F_{n+3} - F_{n-1}$, thus \begin{equation}\boxed{ \sum_{x=1}^{1000} F_xF_{x+2017}= \frac{F_{4020} + F_{2016} - F_{4016} - F_{2020}}{5}.} \end{equation} Here is a numerical check using a CAS. So $a+b+c+d+e=4020+2016+4016+2020+5=12077$ (where $e=5<10$).
