Calculate $\sum_{i=2}^{∞} i\,\frac{\operatorname{fib}(i-1)}{2^i}$

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

Hint:

Using Base case in the Binet formula (Proof by strong induction) if $a,b$ are the roots of $$t^2-t-1=0$$

$$S=\sum_{i=2}^\infty\dfrac{i(a^{i-1}-b^{i-1})}{2^i(a-b)}=\dfrac1{a-b}\left(\dfrac1a\sum_{i=2}^\infty i\left(\dfrac a2\right)^i-\dfrac1b\sum_{i=2}^\infty i\left(\dfrac b2\right)^i\right)$$

Use Sum of a power series $n x^n$

Let $f_0=0$, $f_1=1$, and $f_{n+1}=f_{n-1}+f_n$ for $n\geq 2$. Let $$P(x)=\sum_{i=2}^\infty f_{i-1}x^{i}$$ Then $$(1+x)P(x)=\sum_{i=2}^\infty (f_{i-1}+f_{i-2})x^i=\sum_{i=2}^\infty f_ix^i=\frac{1}{x}(P(x)-x^2).$$ It follows that $$P(x)=\frac{x^2}{1-x-x^2}.$$ Hence $$P'(x)=\frac{2x-x^2}{(1-x-x^2)^2}.$$ Differentiating $P(x)$ directly gives $$P'(x)=\sum_{i=2}^\infty if_{i-1} x^{i-1}.$$ Setting the two expressions of $P'(x)$ equal to each other for $x=1/2$ gives $$\sum_{i=1}^\infty i \frac{f_{i-1}}{2^{i-1}}=12.$$ Dividing by 2 gives the answer 6.