Verified answer

By using the definition of the series. It's a Taylor series where a=0:

http://en.wikipedia.org/wiki/Taylor_series#Definit...

Notice that the first few derivatives are

f ' (x) = e^x - 2e^-2x

f ' ' (x) = e^x + 4e^-2x

f ' ' ' (x) = e^x - 8e^-2x

etc.

In general the nth derivative is e^x + (-2)^n e^-2x. Also, f(0) = e^0 + 2e^0 = 1+2 = 3.

So the series is 3 + ((e^x - 2e^-2x)x/1!) + ((e^x + 4e^-2x)x^2 / 2!) + ... + (e^x + ((-2)^n e^-2x)x^n / n!) + ...