The Maclaurin series of #f_{(x)}=e^{-2x}# is
#f_{(x)}=1+(-2x)+(-2x)^2/{2!}+(-2x)^3/{3!}+ . . .#
First Solution Method: The Maclaurin Series of #y=e^z# is
#y=1+z+z^2/{2!}+z^3/{3!}+z^4/{4!}+ . . .#
Let #z=-2x#.
Then #\quad f_{(x)}=e^{-2x}=e^z\quad# and #f_{(x)}# has the same Maclaurin series as the one above except we set #z=-2x# and get
#f_{(x)}=1+(-2x)+(-2x)^2/{2!}+(-2x)^3/{3!}+ . . .#
I used the well known Maclaurin series for #y=e^z# to get the answer. If this series has not been discussed in class, you should use the general definition of a Maclaurin series to get the answer.
The Maclaurin series of #f_{(x)}# is
# f_{(x)}= f_((x=0))## \quad +{f'_((x=0))}/{1!}x#
#\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +{f''_((x=0))}/{2!]x^2#
#\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad+ {f'''_((x=0))}/{3!}x^3+. . . #
^Couldn't put all terms on same line, sorry for poor formatting.
Anyways, the first term is #f_{(x=0)}#. Here, #\quad f_{(x=0)}=e^{-2(0)}=1#.
The second term is #{f'_{(x=0)}}/{1!}x={-2e^{-2(0)}}/1x=-2x#
The third term is #{f''_{(x=0)}}/{2!}x^2={(-2)^2e^{-2(0)}}/{2!}x^2={(-2x)^2}/{2!}#
These are the same terms as in the Maclaurin series I wrote above.
By observing a pattern, the #n^{th}# term of the series is #(-2x)^n/{n!}#
Using a summation sign, the Maclaurin series of #f_{(x)}# can be written instead as
#f_{(x)}=\Sigma_{n=0}^{n=\infty} [(-2x)^n/{n!} ]#
