#### Answer

We would prefer to have $1$ ¢ today, $2$ ¢ tomorrow, $4$ ¢ on day 3, $8$ ¢ on day 4, $16$ ¢ on day 5, and so on, for 30 days.

#### Work Step by Step

Consider having money in a geometric series for 30 days with ${{a}_{1}}=1$ and $r=2$, and $n=30$.
Thus
$\begin{align}
& {{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)} \\
& =\frac{1\left( 1-{{2}^{30}} \right)}{\left( 1-2 \right)} \\
& =\frac{1\left( -1073741823 \right)}{\left( 1-2 \right)} \\
& =1,073,741,823
\end{align}$
Which is much greater than the cost of a brand new BMW and 10 million dollars. We we would rather have $1$ ¢ today, $2$ ¢ tomorrow, $4$ ¢ on day 3, $8$ ¢ on day 4, $16$ ¢ on day 5, and so on, for 30 days.