Summation Of Product Of Fibonacci Numbers Im trying to find out a general term for the following summation of products of fibonacci numbers:--  
$$\sum_{k=4}^{n+1} F_{k}F_{n+5-k}\; , n \geq 3$$
I tried using Binet's equation but I am getting stuck at a certain point. So, I would be very glad if someone could post an answer to my question with a detailed explanation.
Here are the first few values of the summation for different values of n :--
n = 3 , ans = 9
n = 4 , ans = 30
n = 5 , ans = 73
n = 6 , ans = 158  
Note : I have used the usual fibonacci notation. i.e
$$
F_0=0,\;F_1=1,\;F_2=1,\;F_3=2,\;...etc
$$
EDIT
After reading the comments for this question I tried solving it to form a recurrence relation and this is what i ended up with :-- 
$$  
\begin{align*}
G(n)&=\sum_{k=4}^{n+1} F_kF_{n+5-k}\; , n \geq 3\\
G(n)-G(n-1)&=\sum_{k=4}^{n+1} F_kF_{n+5-k}-\sum_{k=4}^{n} F_kF_{n+4-k}\\
&=F_{n+1}F_{4}+\sum_{k=4}^{n}F_kF_{n+3-k}\\
&=F_{n+1}F_{4}+F_{n}F_{3}+\sum_{k=4}^{n-1}F_kF_{n+3-k}\\
\\
&=F_{n+1}F_{4}+F_{n}F_{3}+G(n-2)\\
\\
G(n)&=G(n-1)+G(n-2)+F_{n+1}F_{4}+F_{n}F_{3}\\
\\
\end{align*}
$$  
Is this correct ? And how do I reduce it further ?
 A: Since there are a number of questions very similar to this one, I will answer a more general question which should answer most of them.

Using the closed form for the Fibonacci numbers,
$$
F_n=\frac{\phi^n-(-1/\phi)^n}{\sqrt{5}}\tag{1}
$$
and Lucas numbers
$$
L_n=\phi^n+(-1/\phi)^n\tag{2}
$$
we get
$$
\begin{align}
F_iF_{n-i}
&=\frac{\phi^i-(-1/\phi)^i}{\sqrt{5}}\frac{\phi^{n-i}-(-1/\phi)^{n-i}}{\sqrt{5}}\\
&=\frac{\phi^n+(-1/\phi)^n-(-1)^i\left(\phi^{n-2i}+(-1/\phi)^{n-2i}\right)}{5}\\
&=\frac{L_{n}-(-1)^iL_{n-2i}}{5}\tag{3}
\end{align}
$$
To sum Lucas numbers, we use $(2)$ and the formula for the sum of a geometric series: 
$$
\begin{align}
\sum_{i=j}^{k-1}F_iF_{n-i}
&=\sum_{i=j}^{k-1}\frac{L_{n+5}-(-1)^iL_{n-2i}}{5}\\
&=\color{#C00000}{\frac{k-j}{5}L_{n}}\\
&\color{#00A000}{-\frac15\sum_{i=j}^{k-1}\phi^{n}\left(-1/\phi^2\right)^i}\\
&\color{#0000FF}{-\frac15\sum_{i=j}^{k-1}(-1/\phi)^{n}\left(-\phi^2\right)^i}\\
&=\color{#C00000}{\frac{k-j}{5}L_{n}}\\
&\color{#00A000}{-\frac{\phi^{n}}{5}\frac{\left(-1/\phi^2\right)^k-\left(-1/\phi^2\right)^j}{\left(-1/\phi^2\right)-1}}\\
&\color{#0000FF}{-\frac{(-1/\phi)^{n}}{5}\frac{\left(-\phi^2\right)^k-\left(-\phi^2\right)^j}{\left(-\phi^2\right)-1}}\\
&=\color{#C00000}{\frac{k-j}{5}L_{n}}\\
&\color{#00A000}{+\frac15\frac{-(-1)^{n-k}(-1/\phi)^{2k-n-1}-(-1)^{j}\phi^{n-2j+1}}{\phi+1/\phi}}\\
&\color{#0000FF}{+\frac15\frac{+(-1)^{n-k}\phi^{2k-n-1}+(-1)^{j}(-1/\phi)^{n-2j+1}}{\phi+1/\phi}}\\
&=\frac{k-j}{5}L_{n}+\frac15\left((-1)^{n-k}F_{2k-n-1}-(-1)^jF_{n-2j+1}\right)\tag{4}\\
&=\frac{k-j}{5}(F_{n-1}+F_{n+1})+\frac15\left((-1)^{n-k}F_{2k-n-1}-(-1)^jF_{n-2j+1}\right)\tag{5}
\end{align}
$$

Applying $(4)$ to the current problem yields
$$
\sum_{k=4}^{n+1}F_kF_{n+5-k}=\frac{n-2}{5}(F_{n+6}+F_{n+4})-\frac25F_{n-2}\tag{5}
$$
