<PRE><B>Question 3978</B>
The partial sum of the first n terms of a geometric series whose first term is a1 and common ratio is r (not = 1) is given by:

Sn = a1(1-r^n)/(1-r)  

But we need to find the common ratio, r.
We can get it from:  an = a1r^(n-1)
                     a4 = a1r^(n-1) Substitute a1 = 4, a4 = 256, and n = 4     
                    256 = 4r^(4-1)
                    256 = 4r^3      Divide both sides by 4.
                     64 = r^3       Take the cube root of both sides.
                      r = 4         The common ratio. 

In this geometric series, a1 = 4, a4 = 256, so the series looks like this:

4, 4^2, 4^3, 4^4, ... or 4, 16, 64, 256, ...  and the common ratio, r,  is 4

The partial sum of the first 4 terms S4, is:
 S4 = 4(1 - 4^4)/(1-4)
 S4 = 4(1 - 256)/(-3)
 S4 = 4(-255)/(-3)
 S4 = -1020/(-3)
 S4 = 340   </PRE>