Wrecking Ball

Using the Infinite Geometric Series, $a_{1}$ = 32 and the common ratio is $\frac{1}{2}$

Simplifying, we have $S_{n}$ = $\frac{a_{1}}{1- r}$ = $\frac{32}{1- \frac{1}{2}}$ = 64

Hence the answer is $\boxed{64}$

$32,16,8,...$ $\implies$ Observed that it is a geometric progression with a common ratio of $\dfrac{8}{16}=\dfrac{16}{32}=0.5$

In here we are finding the sum of the geometric progression with infinitely many terms. If $S$ represents this sum then

$S=\dfrac{a_1}{1-r}$ where $a_1$ is the first term and $r$ is the common ratio. Substituting, we have

$S=\dfrac{32}{1-0.5}=\boxed{64}$

Here we see that the terms for distance are decreasing by 2.

They will continue to decrease from $\frac {1}{2}$ to $\frac {1}{4}$ and so on.

Therefore it is an infinite geometric progression.

So a = 32 and r = $\frac {1}{2}$

Therefore $S_n = \frac {a}{1-r} = \frac {32}{0.5} = 64$

So the answer is $\boxed {64}$