Answer
\[\begin{align}
& \left. a \right)1+3x+\frac{9}{2}{{x}^{2}} \\
& \left. b \right)\sum\limits_{k=0}^{\infty }{\frac{{{\left( 3x \right)}^{k}}}{k!}} \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)={{e}^{3x}},\text{ }a=0 \\
& \text{Using the definition of Taylor series for a function}\left( page\,685 \right) \\
& \sum\limits_{k=0}^{\infty }{\frac{{{f}^{\left( k \right)}}\left( a \right)}{k!}{{\left( x-a \right)}^{k}}}\text{ } \\
& =f\left( a \right)+f'\left( a \right)\left( x-a \right)+\frac{f''\left( a \right)}{2!}{{\left( x-a \right)}^{2}}+\cdots \left( \mathbf{1} \right) \\
& \text{Calculate some derivatives and their value at }x=a=0 \\
& f\left( x \right)={{e}^{3x}}\to f\left( 0 \right)={{e}^{0}}=1 \\
& f'\left( x \right)=3{{e}^{3x}}\to f'\left( 0 \right)=3{{e}^{3\left( 0 \right)}}=3 \\
& f''\left( x \right)=9{{e}^{3x}}\to f''\left( 0 \right)=9{{e}^{3\left( 0 \right)}}=9 \\
& f'''\left( x \right)=27{{e}^{3x}}\to f''\left( 0 \right)=27{{e}^{3\left( 0 \right)}}=27 \\
& \\
& \left. a \right)\text{ Substituting the previous result into the formula }\left( \mathbf{1} \right). \\
& \left( \text{Using the nonzero coefficients} \right) \\
& =\frac{1}{0!}{{\left( x-0 \right)}^{0}}+\frac{3}{1!}{{\left( x-0 \right)}^{1}}+\frac{9}{2!}{{\left( x-0 \right)}^{2}} \\
& \text{Simplifying} \\
& =1+3x+\frac{9}{2}{{x}^{2}} \\
& \\
& \left. b \right)\text{ Rewrite the power series} \\
& =\frac{{{3}^{0}}}{0!}{{x}^{0}}+\frac{{{3}^{1}}}{1!}x+\frac{{{3}^{2}}}{2!}{{x}^{2}}+\frac{{{3}^{3}}}{3!}{{x}^{3}} \\
& \text{Using summation notation}\text{, we obtain} \\
& =\sum\limits_{k=0}^{\infty }{\frac{{{\left( 3x \right)}^{k}}}{k!}} \\
\end{align}\]
