This is a geometric series that looks like; [tex]3+3(3)+3(3)^2+...+3(3)^7[/tex] with [tex]a=3[/tex] and [tex]r=3[/tex] And the sum of any geometric series =[tex]\frac{a(1-r^{n+1} )}{1-r}[/tex] where n is the highest power, which is 7 in this case. so, The sum of this series=[tex] \frac{3(1-3^8)}{1-3}= 9840[/tex] Hope this helps!
