16. $5240 \left( 3 - \frac { n + 3 } { 2 ^ { n } } \right)$

Solution: According to the pattern, for a given $n$, folding $n$ times produces figures with dimensions of the form $\left( \frac { 20 } { 2 ^ { k } } \right) \text{ dm} \times \left( \frac { 10 } { 2 ^ { k } } \right) \text{ dm}$ for $k = 0, 1, \cdots, n$. The number of different sizes is $n + 1$. When $n = 4$, there are 5 different sizes. The area of each size is $S _ { n } = \frac { 240 ( n + 1 ) } { 2 ^ { n } }$. Therefore,

$$\begin{gathered}
\sum _ { k = 1 } ^ { n } S _ { k } = 240 \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } = 240 \left( 2 \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } - \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } \right) \\
= 240 \left( \sum _ { k = 0 } ^ { n - 1 } \frac { k + 2 } { 2 ^ { k } } - \sum _ { k = 1 } ^ { n } \frac { k + 1 } { 2 ^ { k } } \right) = 240 \left( 2 - \frac { n + 1 } { 2 ^ { n } } + \sum _ { k = 1 } ^ { n - 1 } \frac { 1 } { 2 ^ { k } } \right) \\
= 240 \left( 3 - \frac { n + 3 } { 2 ^ { n } } \right) \left( \text{dm} ^ { 2 } \right)
\end{gathered}$$

\section*{IV. Solution Questions}