Answer
\[\begin{align}
& \left. a \right)1-\left( x-1 \right)+{{\left( x-1 \right)}^{2}} \\
& \left. b \right)\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}{{\left( x-1 \right)}^{k}}} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{1}{x},\text{ }a=1 \\
& \text{Using the definition of Taylor series for a function}\left( page\,685 \right) \\
& \sum\limits_{k=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}\text{ }\left( \mathbf{1} \right) \\
& =f\left( a \right)+f'\left( a \right)\left( x-a \right)+\frac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\cdots \\
& \text{Calculate some derivatives and their value at }x=1 \\
& f\left( x \right)=\frac{1}{x}\to f\left( 1 \right)=\frac{1}{1}=1 \\
& f'\left( x \right)=-\frac{1}{{{x}^{2}}}\to f'\left( 1 \right)=-\frac{1}{{{\left( 1 \right)}^{2}}}=-1 \\
& f''\left( x \right)=2{{x}^{-3}}\to f''\left( 1 \right)=2{{\left( 1 \right)}^{-3}}=2 \\
& \\
& \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\
& \left( \text{Using the nonzero coefficients} \right) \\
& =\frac{1}{0!}{{\left( x-1 \right)}^{0}}+\frac{-1}{1!}{{\left( x-1 \right)}^{1}}+\frac{2}{2!}{{\left( x-1 \right)}^{2}} \\
& \text{Simplifying} \\
& =1-\left( x-1 \right)+{{\left( x-1 \right)}^{2}} \\
& \\
& \left. b \right)\text{ Rewrite the power series} \\
& ={{\left( -1 \right)}^{0}}{{\left( x-1 \right)}^{0}}+{{\left( -1 \right)}^{1}}{{\left( x-1 \right)}^{1}}+{{\left( -1 \right)}^{2}}{{\left( x-1 \right)}^{2}} \\
& \text{Using summation notation}\text{, we obtain} \\
& \sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}{{\left( x-1 \right)}^{k}}} \\
\end{align}\]
