When do I multiply that 2 back in, do I just put it in front of the e or does it get multiplied back into the argument before I take the e?
or is it
So sum is:
Also, am I supposed to have an x in my answer? That doesn't seem like an "answer".
No the sum of the series is period.
Since sum(x^n/n!) = e^x and you are substituting (-2x^6) for x
The sum is a Taylor Series so it better have an x in it.
What it can't have is n which is what you're summing over.
not sure where you're getting 1/x^6 or 1/(4*x^6) from but they have nothing to do with the series.