Homework$-9$: Problem $11$

$(8$ points$)$

Match the series with the right expression. $($Use the Maclaurin series.$)$

${\displaystyle \underset{n=0}{\overset{}{}}}\frac{(-1{)}^n{3}^{2n}}{(2n)!}$

${\displaystyle \underset{n=0}{\overset{}{}}}(-1{)}^n\frac{3^{2n+1}}{(2n+1)!}$

${\displaystyle \underset{n=0}{\overset{}{}}}\frac{3^n}{n!}$

${\displaystyle \underset{n=0}{\overset{}{}}}\frac{(-1{)}^n{3}^{-(2n+1)}}{2n+1}$

A$.arctan(\frac{1}{3})$

B$.e^3$

C$.cos(3)$

D$.sin(3)$