#### Answer

For $5$ years, the company B salary is $\$780$ more than company A.

#### Work Step by Step

From the given information, we can observe that it is a geometric series,
For company A:
$30000,1\cdot 06\left( 30000 \right),1\cdot {{06}^{2}}\left( 30000 \right),\cdots $
Here ${{a}_{1}}=30000$ and common ratio $r=1\cdot 06$ and $n=5$.
We use the formula for the sum of a geometric series with finite terms ${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$.
So;
$\begin{align}
& {{S}_{n}}=\frac{30000\left( 1-1\cdot {{06}^{5}} \right)}{\left( 1-1\cdot 06 \right)} \\
& {{S}_{n}}=\frac{30000\left( 1-1\cdot 33825578 \right)}{-\left( 0\cdot 06 \right)} \\
& {{S}_{n}}=\frac{30000\left( -0\cdot 33825578 \right)}{-\left( 0\cdot 06 \right)} \\
& {{S}_{n}}=169,112.77 \\
\end{align}$
For company A, the total salary for 5 years is $169112.77$.
For company B:
$32000,1\cdot 03\left( 32000 \right),1\cdot {{03}^{2}}\left( 32000 \right),\cdots $
Here ${{a}_{1}}=32000$ and common ratio $r=1\cdot 03$ and $n=5$.
We use the formula for the sum of a geometric series with finite terms ${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$.
So;
$\begin{align}
& {{S}_{n}}=\frac{32000\left( 1-1\cdot {{03}^{5}} \right)}{\left( 1-1\cdot 03 \right)} \\
& {{S}_{n}}=\frac{32000\left( 1-1\cdot 1592 \right)}{-\left( 0\cdot 03 \right)} \\
& {{S}_{n}}=\frac{32000\left( -0\cdot 1592 \right)}{-\left( 0\cdot 03 \right)} \\
& {{S}_{n}}=169892.35 \\
\end{align}$
For company B, the total salary for 5 years is $169892.35$.
The difference of total salary for both companies is:
$169112-169892=-780$