# How to find the generating function of one recurrence relation in terms of that of another

I've just started learning about generating functions and recurrence relations and came across this question in my book:

Find the generating function of the recurrence relation $$b_n = (−1)^n(n+1)a_0 + (−1)^{n-1}n a_1 + · · · +(−1)2a_{n-1} + a_n$$

I've put this into a summation, but that's pretty much it.

Any help on how to proceed with this question will be greatly appreciated! It's eating me up that I can't seem to solve this :(

Lets call the generating function of the $a_n$ in the following way $$A = \sum _{n = 0}^{\infty }a_nx^n,$$ then $$-A*(\sum _{m = 0}^{\infty }(-1)^mmx^m) = \sum _{n = 0}^{\infty }x^n\left (\sum _{i+j=n}(-1)^{i+1}ia_j\right )=\sum _{n = 0}^{\infty }x^n\left (\sum _{i=0}^{n}(-1)^{i+1}ia_{n-i}\right ),$$ where the second step is the Cauchy multiplication of series.
So the $n+1-$th coefficient of this new series is $a_n-2a_{n-1}+3a_{n-2}+\cdots +(-1)^{n+2}(n+1)a_0,$
Now, $\sum _{m=0}^{\infty}(-1)^mmx^m=\sum _{m = 0}^{\infty}m(-x)^m=-x\sum _{m = 1}^{\infty }m(-x)^{m-1}=x(\sum _{m = 0}^{\infty}(-x)^m)'=x(\frac{1}{1+x})',$