Arithmetic-geometric series which includes Fibonacci In connection with a problem I'm solving, I seem to be getting the series $$S = 4 \cdot \frac{F_1}{4}+5 \cdot \frac{F_2}{8}+6 \cdot \frac{F_3}{16}+7 \cdot \frac{F_4}{32}+ \cdots$$
where $F_i$ are the Fibonacci numbers. How does one solve this?
I can see here that the Fibonacci sequence is related to a geometric sequence, but I'm not sure how to incorporate that. 
 A: Let $$F(x) =\sum _{i=0}^{\infty}F_nx^n$$ Then by Infinite Series: Fibonacci/ $2^n$
we have 
$$F(x) = {x\over 1-x-x^2}$$
we are interested in $F'(1/2)$ so ... 
A: Consider the generating function $$F(x)=\sum_{n=1}^{\infty}F_nx^n$$  If we index the Fibonacci numbers as $F_1=1=F_2$ then this can be computed as $$F(x)=\frac x{1-(x+x^2)}$$
See, e.g., this question
We now just have to adapt this to suit your series.
First remark that $$G(x)=x^3F(x)=\frac {x^4}{1-(x+x^2)}=\sum_{n=1}^{\infty}F_nx^{n+3}$$
It follows that $$\frac d{dx}G(x)=3x^2F(x)+x^3F'(x)=\sum_{n=1}^{\infty}(n+3)F_nx^{n+2}$$
Taking $x=\frac 12$ yields the sum $4\frac {F_1}8+5 \frac {F_2}{16}+\cdots$ which is eaxactly half of your sum.  Thus the answer you seek is $$2\times \left(3\times \left(\frac 12\right)^2\times F\left(\frac 12\right)+\left(\frac 12\right)^3F'\left(\frac 12\right)\right)$$
This is (relatively) easy to compute and yields $$\boxed 8$$
Worth remarking:  This agrees with numerical computation (summing the first hundred terms yields $7.999999936$).
