Which is the correct taylor series

Question

Find the Taylor series for f(x)=2xe−3x at x=4 and its interval of convergence.

∑n=0∞1/n! e^-12(8(−3)ⁿ−2n(−3)^n-1)(x−4)ⁿ,(−∞,∞)

∑n=0∞1/n! e^-12(8(−3)ⁿ+2n(−3)^n-1)(x−4)ⁿ,(−∞,∞)

∑n=0∞1/n! e^-122n(−3)^n-1(x−4)ⁿ,(−∞,∞)

∑n=0∞1/n! e^-128n(−3)ⁿ(x−4)ⁿ,(−∞,∞)

Answer 1

f(x)=2xe^-3x --> f(4) =8e^-12

f'(x)=2e^-3x-6xe^-3x --> f'(4) = -22e^-12

f''(x) = -6e^-3x-6e^-3x+18xe^-1x --> f"(4) =-6e^-12-6e^-12+72e^-12 = 60e^-12

^...

Looking at only the coefficients for e^-12 -->8, -22, 60,...

8(-3)ⁿ+2n(-3)^n-1 fits for n = 0,1,2

So the second choice is the answer.