Answer
$S_{9}=10220$
Work Step by Step
We are asked to find the sum of:
$5120+2560+1280+...+20$
We see that this is a geometric sequence with $a_1=5120$. We find $r$:
$r=2560/5120=0.5$
We know that a geometric sequence has the form:
$a_{n}=ar^{n-1}$
We use this to find $n$:
$a_n=(5120)(0.5)^{n-1}$
$20=(5120)(0.5)^{n-1}$
$20=5120*(0.5)^n/0.5$
$10=5120*(0.5)^n$
$\frac{1}{512}=(\frac{1}{2})^n$
$512=2^n$
$n=\log_2 512=9$
We know the partial sum of a geometric sequence is:
$S_n=a_1\frac{1-r^n}{1-r}$
$S_{9}=5120 \frac{1-(\frac{1}{2})^{9}}{1-\frac{1}{2}}=10220$
