Find the sum of the progression 8, 4, 2, ..., 1/512 leaving the answer to the nearest whole number.

The given progression is $8, 4, 2, \dots, \frac{1}{512}$.
This is a geometric progression (GP) because each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
The first term is $a = 8$.
The common ratio is $r = \frac{4}{8} = \frac{1}{2}$.
The last term is $T_n = \frac{1}{512}$.

The formula for the $n$-th term of a geometric progression is $T_n = ar^{n-1}$.
Substitute the known values:

Divide both sides by 8:

Since $2^{12} = 4096$, we can write $\frac{1}{4096}$ as $\left(\frac{1}{2}\right)^{12}$:

Equating the exponents:

There are 13 terms in the progression.

The formula for the sum of the first $n$ terms of a geometric progression when $r < 1$ is $S_n = \frac{a(1-r^n)}{1-r}$.
Substitute $a=8$, $r=\frac{1}{2}$, and $n=13$:

Calculate $\left(\frac{1}{2}\right)^{13}$:

Substitute this back into the sum formula:

Multiply the numerator by 2 (which is the reciprocal of $\frac{1}{2}$):

Rounding to the nearest whole number, 15.998... becomes 16.