SOLUTION: What is Sn of the geometric series with a1 = 4, an = 256 and n = 4?

Question 3978: What is Sn of the geometric series with a1 = 4, an = 256 and n = 4?
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
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
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