Summation of series of product of Fibonacci numbers What is the sum of following product of Fibonacci numbers 
$$\sum_{k=1}^{n-1} Fib(k)*Fib(n+3-k)$$
can anyone suggest only approach to find general term?
 A: Assuming you're using the usual numbering of the Fibonacci numbers, the Binet equation
$$\text{fib}(n) = \frac{\phi^n - (-1/\phi)^n}{\sqrt{5}}$$
where $\phi = (1 + \sqrt{5})/2$ implies your sum is
$$ F(n) = \left((2 + \sqrt{5}) \frac{n}{5} - \frac{7 \sqrt{5}}{25}\right) \phi^n + 
\left((2 - \sqrt{5}) \frac{n}{5} + \frac{7 \sqrt{5}}{25}\right) (-1/\phi)^n
$$
The first few terms ($n = 1$ to $10$) are $0, 3, 8, 19, 40, 80, 154, 289, 532, 965$.
EDIT: It seems your $\text{Fib}(n) = \text{fib}(n+1)$.  So your sum is
$$\sum_{k=1}^{n-1} \text{fib}(k+1) \text{fib}(n+4-k) = \left(\frac{\sqrt {5} n}{2}+{\frac {11 n}{10}}-{\frac {21}{50}}\sqrt {5}-\frac{3}{2} \right) \phi^n + \left(-\frac{\sqrt {5} n}{2}+{\frac {11 n}{10}}+{\frac {21}{50}}\sqrt {5}-\frac{3}{2} \right) (-1/\phi)^n
$$
The first few terms ($n=1$ to $10$) are $0, 5, 18, 44, 96, 195, 380, 719, 1332, 2428$.
The generating function is
$$g(t) = {\frac {t \left( 3\,t+5 \right)  \left( t+1 \right) }{ \left( {t}^{2}+
t-1 \right) ^{2}}}
$$
A: It is also possible to compute function values one at a time by using the recursion formula $G(n) = 2G(n-1)+G(n-2)-2G(n-3)-G(n-4)$ after manually computing the first four values. (The validity of the formula follows e.g. with Robort Israel's explicit formula and the defining identity for $\phi$; note that $x^4-(2x^3+x^2-2x-1)=(x^2-x-1)^2$).
