How to prove this Fibonacci identity? $\sum_{k=0}^{n} F_{k} F_{n-k} = \frac{1}{5}\left(n L_{n} - F_{n}\right)$ How to prove this Fibonacci identity?
$$\sum_{k=0}^{n-3} F_{k} F_{n-k-3} = \frac{(n-3)L_{n-3} - F_{n-3}}{5}$$
 i tried to used the generating function and partial decomposition but i got confused?

Using 
\begin{align}
F_{n} &= \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta} \\
L_{n} &= \alpha^{n} + \beta^{n} \\
2 \alpha &= 1 + \sqrt{5} \\
2 \beta &= 1 - \sqrt{5}
\end{align}
then
\begin{align}
(\alpha - \beta)^{2} \, F_{n} F_{n-k} &= L_{n} - \beta^{n} \, \left(\frac{\alpha}{\beta}\right)^{k} - \alpha^{n} \, \left(\frac{\beta}{\alpha}\right)^{k} \\
&= L_{n} - \alpha^{n} (-\beta^{2})^{k} - \beta^{n} (- \alpha^{2})^{k}.
\end{align}
Now,
\begin{align}
5 \, \sum_{k=1}^{n} F_{k} F_{n-k} &= n \, L_{n} - \alpha^{n} \cdot \frac{(- \beta^{2}) ( 1- (-1)^{n} \beta^{2n})}{1 + \beta^{2}} - \beta^{n} \cdot \frac{(- \alpha^{2}) ( 1- (-1)^{n} \alpha^{2n})}{1 + \alpha^{2}} \\
&= n \, L_{n} - \beta \, F_{n} - \alpha \, F_{n} \\
&= n \, L_{n} - F_{n}.  
\end{align}
Since $F_{0} = 0$ then
$$\sum_{k=0}^{n} F_{k} \, F_{n-k} = \frac{n \, L_{n} - F_{n}}{5}.$$
