We will have that it is solved as follows:
[tex]\sum ^9_{k=1}256(\frac{1}{2})^{k-1}=256+128+64+32+16+8+4+2+1=511[/tex]***
To solve we replace the values in the formula given:
[tex]S_n=a_1(\frac{1-r^n}{1-r})[/tex]Now, we have to determine the ratio, which is 1/2, and since we want to know the value at n = 9 and the fist value of the series (a1) is 256, we replace it as well:
[tex]S_9=256(\frac{1-(\frac{1}{2})^9}{1-\frac{1}{2}})\Rightarrow S_n=511[/tex]