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?
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?
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}}} $
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 $x^4-(2x^3+x^2-2x-1)=(x^2-x-1)^2$).