Chapter 9.10, Problem 4QCE

Suppose that $f(1)=4\text{and}f\text{'}(x)={\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{{(-1)}^k}{(k+1)!}{(x-1)}^k}$

$\begin{array}{l}\text{(a)}f"(1)=\text{\_\_\_\_\_\_}\\ \left(\text{b}\right)f(x)=\text{\_\_\_\_\_+\_\_\_\_\_(}x-\text{1)+\_\_\_\_\_(}x-1{)}^2+\text{\_\_\_\_\_\_\_(}x-1{)}^3+\mathrm{...}=\text{\_\_\_\_\_}+{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\text{\_\_\_\_}}\end{array}$