How to simplify coefficient of Maclaurin series

Question

I have the following exponential generating function G(x),

\begin{align*} {G(x)}&=\left(e^x-1\right)(e^x-1-x)e^x\\ &{=e^{3x}\color{blue}{-(x+2)e^{2x}+(x+1)e^x}} \end{align*}

As far as I see, I can write $e^{3x}$ as $3^n$ although I don't know the logic behind it.. But I don't know how to write the blue part as such.

How is the following obtained, for example, $n2^{n-1}$ and the others? Could you please help and explain?

$$\sum_{n=1}^{\infty} a_{n} \frac{x^n}{n!} = \sum_{n=1}^{\infty} 3^{n} \frac{x^n}{n!} - \sum_{n=1}^{\infty}n2^{n-1} \frac{x^n}{n!} - \sum_{n=1}^{\infty} 2^{n+1} \frac{x^n}{n!} + \sum_{n=1}^{\infty}(n+1) \frac{x^n}{n!}$$

Answer 1

It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.

We obtain for $n\geq 1$: \begin{align*} \color{blue}{[x^n]G(x)}&=[x^n]\left(e^{3x}-(x+2)e^{2x}+(x+1)e^x\right)\\ &=[x^n]e^{3x}-[x^n](x+2)e^{2x}+[x^n](x+1)e^x\tag{1}\\ &=[x^n]e^{3x}-\left([x^{n-1}]+2[x^n]\right)e^{2x}+\left([x^{n-1}]+[x^n]\right)e^x\tag{2}\\ &=[x^n]\sum_{k=0}^{\infty}\frac{(3x)^k}{k!}-\left([x^{n-1}]+2[x^n]\right)\sum_{k=0}^{\infty}\frac{(2x)^k}{k!}\\ &\qquad+\left([x^{n-1}]+[x^n]\right)\sum_{k=0}^{\infty}\frac{x^k}{k!}\tag{3}\\ &=\frac{3^n}{n!}-\left(\frac{2^{n-1}}{(n-1)!}+2\frac{2^n}{n!}\right)+\left(\frac{1}{(n-1)!}+\frac{1}{n!}\right)\tag{4}\\ &\,\,\color{blue}{=\frac{1}{n!}\left(3^n-n2^{n-1}-2^{n+1}+n+1\right)}\tag{5} \end{align*}

Comment:

• In (1) we use the linearity of the coefficient of operator.

• In (2) we apply the rule $[x^{p-q}]A(x)=[x^p]x^qA(x)$.

• In (3) we use the series expansion of the series at $x=0$.

• In (4) we select the coefficient of $[x^n]$ and $[x^{n-1}]$.

Since the constant part of the series is \begin{align*} \color{blue}{[x^0]G(x)}&=[x^n]\left(e^{3x}-(x+2)e^{2x}+(x+1)e^x\right)\\ &=1-(0+2)+(0+1)\\ &\,\,\color{blue}{=0}\tag{6} \end{align*} we obtain from (5) and (6) \begin{align*} \color{blue}{G(x)}&\color{blue}{=\sum_{n=1}^{\infty}\left(3^n-n2^{n-1}-2^{n+1}+n+1\right)\frac{x^n}{n!}} \end{align*} according to the claim.