# Combinatorics-Exponential Generating Function(Kenneth Rosen ex 6.4 47e)

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.

• You are correct. – Robert Z Jun 26 '18 at 7:33
• It is not necessary to mention $a_0=0$, since $(-2)^{0}-0!=1-1=0$. – Foobaz John Jun 26 '18 at 15:23