So i have a function: $$f: \mathbb{R} \to \mathbb{R}$$ $$f(x)=(1-4x^2)e^{2x}$$

I have to formulate it's Taylor Series around x=0 (so i think it is a Maclaurin series if i recall correctly). And then i have to use it as help for calculating sum : $$ \sum_{n=2}^{\infty} \frac{(-2)^n(n^2 -n-1)}{n!}$$

So i know i will have to use the sum of $e^x$ that is a known Maclaurin series. Butother than that, help would be appreciated.


Notice that for $n\ge2$,


so that

$$\sum_{n=2}^\infty\frac{n^2-n-1}{n!}(2x)^n=\sum_{n=2}^\infty(2x)^2\frac{(2x)^{n-2}}{(n-2)!}-\sum_{n=2}^\infty\frac{(2x)^n}{n!}=(2x)^2e^{2x}-\left(e^{2x}-1-2x\right)\\ =(4x^2-1)e^{2x}+1+2x.$$

Then with $x=-1$,



Since the taylor series for $f(x)=(1-4x^2)e^{2x}$ is $$\sum_{n=0}^{1} \frac{2^n}{n!} x^n- \sum_{n=2}^{\infty} \frac{(2)^n(n^2 -n-1)}{n!}x^n$$

Let $x = -1$. $$\sum_{n=0}^{1} \frac{2^n}{n!} (-1)^n- \sum_{n=2}^{\infty} \frac{(2)^n(n^2 -n-1)}{n!}(-1)^n = -1- \sum_{n=2}^{\infty} \frac{(-2)^n(n^2 -n-1)}{n!} = -3/e^2 $$

Then $$\sum_{n=2}^{\infty} \frac{(-2)^n(n^2 -n-1)}{n!} = 3/e^2 -1$$

  • $\begingroup$ How do you get the Taylor series ? $\endgroup$
    – user65203
    Aug 1 '16 at 7:43
  • $\begingroup$ @MathIsTheWayOfLife you are welcome. $\endgroup$
    – Zack Ni
    Aug 1 '16 at 7:45

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