I guess I'm having difficulty with this because its not in the form of a polynomial expression, which is what I've been taught. Nevertheless here's what I did:
I know that the expansion for $e^{2x}=\sum x^{2k}/2k!$. I'm not entirely sure what to do next but I have to basically subtract 1 and 2x and divide by $x^5$. Then I get this expression
$(\sum (2x)^{k}/k!-1-2x)/x^5$
However, I know that Laurent Series have to look something like: $\sum a_n(x-c)^n$, with a sum that goes from negative to positive infinity. I really don't know what to do and could really use some valuable guidance.
Your series for $e^{2x}$ should be $$e^{2x} = \sum_{k=0}^{\infty} \frac{(2x)^{k}}{k!}.$$ Note that this is equal to $$e^{2x} = 1 + 2x + \sum_{k=2}^{\infty} \frac{(2x)^{k}}{k!},$$ so we have that $$e^{2x} - 1 - 2x = \sum_{k=2}^{\infty} \frac{(2x)^{k}}{k!}.$$ Finally, by dividing by $x^{5}$, we get $$\frac{e^{2x}-1-2x}{x^{5}} = \sum_{k=2}^{\infty} \frac{2^{k}}{k!} x^{k-5} = \sum_{k=-3}^{\infty} \frac{2^{k+5}}{(k+5)!} x^{k}.$$ If you insist on putting it in the form $\sum_{k=-\infty}^{\infty} a_{k} (x-c)^{k}$, then we have $c = 0, a_{k} = 0$ for $k < -3$, and $a_{k} = \frac{2^{k+5}}{(k+5)!}$ for $k \geq -3$.