Calculate $\sum_{i=2}^{∞} i\,\frac{\operatorname{fib}(i-1)}{2^i}$ Calculate: $$\sum_{i=2}^{∞} i\,\frac{\operatorname{fib}(i-1)}{2^i}$$ Where $\operatorname{fib}(i)$ is the $i$-th Fibonacci number. I know that it's $6$ (based on a program), but how to prove it?

For $i \ge 2$, you have (wikipedia link)

$$\operatorname{fib}(i-1) = \frac{\phi^{i-1}-\phi^{\prime (i-1)}}{\sqrt 5}$$ where

$$\phi = \frac{1 + \sqrt 5}{2} \text{ and } \phi^\prime = \frac{1 - \sqrt 5}{2}$$

Therefore $$I=\sum_{i=2}^{\infty} i\,\frac{\operatorname{fib}(i-1)}{2^i} = \frac{1}{2\sqrt 5} \left(\sum_{i=2}^{\infty} i\left(\frac{\phi}{2}\right)^{i-1} - \sum_{i=2}^{\infty} i \left(\frac{\phi^\prime}{2}\right)^ {i-1}\right)$$

Now for $0 < x < 1$, $S(x) = \sum_{i=0}^{\infty} x^i = \frac{1}{1-x}$ and $S^\prime(x) = \sum_{i=1}^{\infty} i x^{i-1} = \frac{1}{(1-x)^2}$.

Which leads to $$I = \frac{1}{2\sqrt 5}\left( \frac{1}{(1-\phi/2)^2} - \frac{1}{(1-\phi^\prime/2)^2}\right)=6$$