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Find the sequence with each of these functions as its exponential generating function
g(x) = $e^{-2x} - \frac{1}{1-x}$

so I expanded it like
$\sum_{r=0}^{\infty}\frac{(-2^r)x^r}{r!} \,-\sum_{r=0}^{\infty}x^r$
and finally after resolving it i got expression below

$\left\{ (\frac{-2-1!}{1!}).x + (\frac{2^2-2!}{2!}).x^2  + (\frac{-2^3-3!}{3!}).x^3 + (\frac{2^4-4!}{4!}).x^4+.......\right\}$
so This is my exponential series that I got and it's coefficient term can be generalized as
$a_n=(-2)^n -n!\, , for\, n\,\geq1$ and $a_0=0$
but rosen's key says it's $a_n=(-2)^n+n!$ and they even have not given any condition which implies it is valid for $n\geq0$. So, according to rosen's answer $a_0$ should exist but in my answer $a_0$ is 0. Please let me know where I am missing something.
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$G\left ( x \right )=e^{-2x}-\frac{1}{1-x}$

$G\left ( x \right )=\sum \frac{\left ( -2 \right )^{n}.x^{n}}{n!}-\sum x^{n}$

                  $=\sum \left ( \frac{\left ( -2 \right )^{n}.}{n!}- 1 \right ).x^{n}$

$a_{n}=\left ( \frac{\left ( -2 \right )^{n}.}{n!}- 1 \right )$

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