Question

In Problems 11-16 use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation.

$\mathit{x}{\mathit{e}}^{\mathbf{3}\mathbf{x}}$

Short answer

$x{e}^{3x}=\sum _{n=0}^{\infty }\frac{3^n{x}^{n+1}}{n!}$

Step-by-step solution

Definition

Maclaurin seriesare a type of series expansion in which all terms are nonnegative integer powers of the variable.

Find Maclaurin series

We have Maclaurin series

$$\begin{gathered} e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ =\sum _{n=0}^{\infty }\frac{1}{n!}{x}^n \end{gathered}$$

Now replace$x$ as$3x$ we get$e^{3x}=1+\frac{3x}{1!}+\frac{{\left(3x\right)}^2}{2!}+\frac{{\left(3x\right)}^3}{3!}+\dots ----\left(1\right)$

Now multiply equation (1) with $x$and we get

$$\begin{gathered} x·e^{3x}=x·1+x·\frac{3x}{1!}+x·\frac{{\left(3x\right)}^2}{2!}+x·\frac{{\left(3x\right)}^3}{3!}+\dots \\ =x+\frac{3x^2}{1!}+\frac{3^2{x}^3}{2!}+\frac{3^3{x}^4}{3!}+\dots \\ =\sum _{n=0}^{\infty }\frac{3^n{x}^{n+1}}{n!} \end{gathered}$$

Thus the Maclaurin series for$x{e}^{3x}$ is$\sum _{n=0}^{\infty }\frac{3^n{x}^{n+1}}{n!}$