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**Scott H** Your calculations for the derivatives are correct. The Maclaurin expansion is

$\displaystyle f(x)=f(0)+f'(0)x+f''(0)\frac{x^2}{2!}+f'''(0)\frac {x^3}{3!}+\cdots$

Here, we have

$\displaystyle \begin{aligned}

f(x)&=\frac{1}{2^3}+\frac{(-3)x}{2^4}+\frac{(-3)(-4)x^2}{2^5\cdot2!}+\frac{(-3)(-4)(-5)x^3}{2^6\cdot3!}+\cdots\\

&=\frac{1}{2^3}-\frac{3\cdot x}{2^4}+\frac{3\cdot 4\cdot x^2}{2^5\cdot 2!}-\frac{3\cdot 4\cdot 5\cdot x^3}{2^6\cdot 3!}+\cdots

\end{aligned}

$

Now we look closely at the terms and think of a formula for the $\displaystyle n$th term in terms of $\displaystyle n$ (in our case, starting at $\displaystyle n=0$ will be better).

First, we notice that this is an alternating series. We can start with the factor $\displaystyle (-1)^n$ and then add the factors

$\displaystyle x^n,\,\frac{1}{2^{n+3}},\,\frac{1}{n!},\,3\cdot 4\cdot 5\cdots \cdot (n+1)\cdot(n+2)=\frac{(n+2)!}{2},$

giving us

$\displaystyle \sum_{n=0}^\infty\,(-1)^n\cdot\frac{(n+2)!}{2}\cdot\frac{1}{2^{n+3}}\cd ot x^n\cdot\frac{1}{n!}=\sum_{n=0}^\infty\,(-1)^n\frac{(n+2)!x^n}{2^{n+4}\cdot n!}.$