Find The Maclaurin series of $5x^2e^{-8x}$

I'm having trouble finding the Maclaurin series of $5x^2e^{-8x}$ and determining the coefficients. the initial $e^{-8x}$ is simple enough but after that I get confused, our teacher rushed through Maclaurin series so he didn't really explain the concept very well. The homework gives the answers to the coefficients, C1=0, C2=5, C3=-40 But I can't seem to understand how those answers come from the series. The answer I get is ,

$5\cdot(\sum_{n= 0}(-1)^n\frac{(8)^{n}(x)^{n+2}}{n!})$

• Try writing out the first few terms of your series explicitly. – Michael Seifert Dec 7 '17 at 19:22
• Your series seem to be correct (Mind you, for n=0 your first term is quadratic!) – imranfat Dec 7 '17 at 19:22
• I like your answer, and the first few terms would be $5x^2 - 40 x^3 + 160 x^4$ – Doug M Dec 7 '17 at 19:23
• Okay i'll try, but another issue is how exactly do you calculate the Coefficients, is there a reason why C1=0? – M.Bucciacchio Dec 7 '17 at 19:24
• Yes, the answer is in the exponent of $x$ in your summation – imranfat Dec 7 '17 at 19:24

We have $$e^t=\sum_{n=0}^{\infty}\frac{t^n}{n!}.$$ Put $t=-8x$ and multiply by $5x^2$. So,
$$5x^2e^{-8x}=5x^2\sum_{n=0}^{\infty}\frac{(-8x)^n}{n!}=\sum_{n=0}^{\infty}\frac{(-1)^n\cdot 5\cdot 8^nx^{n+2}}{n!}.$$
Maclaurin series for $e^{x}$ is $$1+x+\dfrac{x^2}{2!}+... =\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$$ so by the Composition Rule (with $x$ replaced by $-8x$), the series becomes $$e^{-8x}=\sum_{n=0}^{\infty}\dfrac{(-8x)^n}{n!}$$ and thus by the Product Rule the desired Maclaurin series is $$5x^2e^{-8x}=5x^2\sum_{n=0}^{\infty}\dfrac{(8x)^n}{n!}=5\sum_{n=0}^{\infty}\dfrac{(-1)^n\cdot 8^n \cdot x^{2n+2}}{n!}$$