Answer
\[\begin{align}
& \left. a \right){{x}^{2}}-{{x}^{3}}+{{x}^{4}} \\
& \left. b \right)\sum\limits_{k=2}^{\infty }{{{\left( -1 \right)}^{k}}{{x}^{k}}} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\frac{{{x}^{2}}}{1+x},\text{ }a=0 \\
& \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{First}\text{, we calculate some derivatives and their value at }x=0 \\
& f\left( x \right)=\frac{{{x}^{2}}}{1+x}\to f\left( 0 \right)=\frac{{{\left( 0 \right)}^{2}}}{1+\left( 0 \right)}=0 \\
& f'\left( x \right)=\frac{2x\left( 1+x \right)-{{x}^{2}}}{{{\left( 1+x \right)}^{2}}}=\frac{2x+{{x}^{2}}}{{{\left( 1+x \right)}^{2}}}\to f'\left( 0 \right)=0 \\
& f''\left( x \right)=\frac{{{\left( 1+x \right)}^{2}}\left( 2+2x \right)-2\left( 1+x \right)\left( 2x+{{x}^{2}} \right)}{{{\left( 1+x \right)}^{4}}} \\
& \text{Simplifying} \\
& f''\left( x \right)=\frac{\left( 1+x \right)\left( 2+2x \right)-2\left( 2x+{{x}^{2}} \right)}{{{\left( 1+x \right)}^{3}}} \\
& f''\left( x \right)=\frac{2}{{{\left( 1+x \right)}^{3}}} \\
& f''\left( 0 \right)=\frac{2}{{{\left( 1+x \right)}^{3}}}\to f''\left( 0 \right)=2 \\
& f'''\left( x \right)=2\left( -3 \right){{\left( 1+x \right)}^{-4}}=-\frac{6}{{{\left( 1+x \right)}^{4}}}\to f'''\left( 0 \right)=-6 \\
& {{f}^{\left( 4 \right)}}\left( x \right)=-6\left( -4 \right){{\left( 1+x \right)}^{-5}}=\frac{24}{{{\left( 1+x \right)}^{5}}}\to {{f}^{\left( 4 \right)}}\left( 0 \right)=24 \\
& \\
& \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\
& \left( \text{Using the nonzero coefficients} \right) \\
& =0+0+\frac{2}{2!}{{x}^{2}}+\frac{\left( -6 \right)}{3!}{{x}^{3}}+\frac{24}{4!}{{x}^{4}} \\
& \text{Simplifying} \\
& ={{x}^{2}}-{{x}^{3}}+{{x}^{4}} \\
& \\
& \left. b \right)\text{ Rewrite the power series} \\
& ={{\left( 1 \right)}^{2}}{{x}^{2}}+{{\left( -1 \right)}^{3}}{{x}^{3}}+{{\left( -1 \right)}^{4}}{{x}^{4}} \\
& \text{Using summation notation}\text{, we obtain} \\
& \sum\limits_{k=2}^{\infty }{{{\left( -1 \right)}^{k}}{{x}^{k}}} \\
\end{align}\]
