How to prove this Fibonacci identity? $\sum_{k=0}^{n} F_{k} F_{n-k} = \frac{1}{5}\left(n L_{n} - F_{n}\right)$

Question

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?

Answer 1

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 $$\sum_{k=0}^{n} F_{k} \, F_{n-k} = \frac{n \, L_{n} - F_{n}}{5}.$$