Calculate $\sum_{k=0}^{n} \frac{F_k}{2^k}$ Hint: use convolution of generating functions
My try:
$a_i = 2^i$ $b_i = F_i$ $$\sum_{k=0}^{n} \frac{F_k}{2^k} = 2^{-n} \sum_{k=0}^{n} F_k 2^{n-k} $$ and now let $$c_n = \sum_{k=0}^{n} F_k 2^{n-k} $$
I know that $<a_n> \bigoplus <b_n> = <c_n>$ So let $$A(x) = \sum_i a_i $$ and similar $B(x)$
$$A(X)B(x) = \sum_{i=0} c_i $$ but I stucked there, how can I finish this task?
Depending on how you index the fibonacci sequence (i.e. is $F_0 = 1$ or $0$?), you'll have, resp., $$B(x) = \sum_i F_i x^i = \frac{1}{1-x-x^2} \text{ or } \frac{x}{1-x-x^2}$$
Since $$A(x) = \sum_i 2^i x^i = \frac{1}{1-2x},$$ we can multiply and use partial fraction decomposition to write $$A(x)B(x) = \frac{k_1x+k_2}{1-x-x^2} + \frac{k_3}{1-2x},$$ (for some constants $k_1,k_2,k_3$ you'll need to find) which you can then use to find a simplified expression for the coefficients of $A(x)B(x).$