Given is:
$f(x)=(3+e^{-x})^2$
so I write:
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$$
$$e^{-x}x=\sum_{n=0}^\infty \frac{(-x)^n}{n!}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots$$
$$f(x)=\left(3+1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\dots\right)^2$$
How do I put $f$($x$) in $\sum$ form?
You have
$$\mathrm{f}(x) = \left(3+\mathrm{e}^{-x}\right)^2 \equiv 9 +6\,\mathrm{e}^{-x}+\mathrm{e}^{-2x}$$
The series are then:
$$9 + 6\left(\sum_{r=0}^{\infty} (-1)^r\frac{x^r}{r!}\right)+\left(\sum_{s=0}^{\infty}(-2)^s\frac{x^s}{s!}\right)$$
$$= 9 + \sum_{t=0}^{\infty}\left[6(-1)^t + (-2)^t\right]\frac{x^t}{t!}$$