Generating function of $\{ a_n \}_{0 \leq n}$ is $$A(x) = \frac{1+2x}{1-2x}$$ then what is the generating function of ${n a_n}$ ?
It is what I tried so far,
$$A(x) = \frac{1+2x}{1-2x}$$ $$= (1+2x)(1-2x)^{-1}$$ $$= (1+2x)(\sum_i \binom{i+1-1}{i}(2x)^i)$$ $$= (1+2x)(1+2x+4x^2+8x^3+...)$$ $$= (\overbrace{1+2x+4x^2+8x^3+...}^{1} + \overbrace{2x+4x^2+16x^3+...}^{2x})$$ $$A(x) = 1+4x+8x^2+24x^3+...$$ So we can say $a_0=1, a_1=4,a_2=8,a_3=24$,...
My problem is:
First: what is the meaning of $\{na_n\}$?, I found every element of $a_n$ is it true to say that $$na_n = $$ $$0 a_0 = 0 \times 1 = 0$$ $$1a_1 = 1 \times 4 = 4$$ $$2a_2 = 2 \times 8 = 16$$ $$...$$ ?
Second, how to calculate $\{na_n\}$ as a formula look like what $A(x)$ is?
Hint: If the generating function of a sequence $\{a_n\}$ is $A(x) = \displaystyle\sum_{n = 0}^{\infty}a_nx^n$, then its derivative is simply $A'(x) = \displaystyle\sum_{n = 0}^{\infty}na_nx^{n-1}$, and so, $xA'(x) = \displaystyle\sum_{n = 0}^{\infty}na_nx^{n}$.