Answer
a. $\approx 99.9$ mg
b. $100$ mg
Work Step by Step
a.
Recognize $S_{n}=\displaystyle \sum_{k=1}^{n}50(\frac{1}{2})^{k-1} $
as the nth partial sum of a geometric sequence $a_{n}=ar^{n-1}$,
where a=50, $r=\displaystyle \frac{1}{2}.$
$S_{n}=a\displaystyle \cdot\frac{1-r^{n}}{1-r}$.
For n=10
$S_{10}=50\displaystyle \cdot\frac{1-(\frac{1}{2})^{10}}{1-\frac{1}{2}}\approx 99.9$ mg
b.
Recognize $\displaystyle \sum_{k=1}^{\infty}50(\frac{1}{2})^{k-1}$
as an infinite geometric series
for which a=50, $r=\displaystyle \frac{1}{2}.$
$|r| < 1$, so the series converges to
$S=\displaystyle \frac{a}{1-r}=\frac{50}{1-\frac{1}{2}}=100$ mg
