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, $\sum _{m=0}^{\infty}(-1)^mmx^m=\sum _{m = 0}^{\infty}m(-x)^m=-x\sum _{m = 1}^{\infty }m(-x)^{m-1}=x(\sum _{m = 0}^{\infty}(-x)^m)'=x(\frac{1}{1+x})',$

Can you finish?


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