If you know the series for e^x, then you can start from there:
e^x = sum (n from 0 to inf) [(x^n) / n!]
=> e^3x = sum (n from 0 to inf) [(3x)^n / n!]
If you don't know the series for e^x, then do it this way:
f(x) = f(0) + x f'(0) + x^2 f''(0) / 2!+ x^3 f'''(0) / 3! + ...
f(0) = e^0 = 1
f'(x) = 3e^3x => f'(0) = 3
f''(x) = (3^2)e^3x => f'(0) = 3^2
f'''(x) = (3^3)e^3x => f'(0) = 3^3
Re-arranging and you'll get the sigma form as above.
