For the geometric series 1 + 3 + 9 + 27 + . . . , find the sum of the first 10 terms.
A. 1,493
B. 4,580
C. 13,841
D. 41,624
I keep doing my math wrong, I cant get the answer.
5 Answers

Your sum is
S = 3^0 + 3^1 + 3^2 + … + 3^9 …. where “^” means “raised to the power”
If you remember the formula for a geometric series, just plug in a=1, r=3 and n=10 and you’re done. If not, then notice what happens when you multiply both sides by 3:
3S = 3^1 + 3^2 + 3^3 + … + 3^9 + 3^10
The last 9 terms in the first sum are the same as the first 9 terms in the 3S sum. Subtract the two and they cancel:
3S – S = 3^10 – 3^0
2S = 3^10 – 1
S = (3^10 – 1) / 2
So, I can’t get the answer either…because it isn’t in the list. 3^10 = 59049, and S = 59048/2 = 29524.

what is the sum of the first 4 terms of the geometric 1. 3. 9 . 27?

The sum of is given by the formula S = a(r^n – 1)/(r – 1).
a = first term = 1
r = common ratio = 3, and
n = 10
So, Sum = 1*(3^10 – 1)/(3 – 1) = 29 524

as in 1+3+9+27+81+243+729+2187+6561+19683?
so far the last number exceeds all but the largest, so i think the answer is obvious now.
as to how to do it i’m pretty sure its got something to do with exponents

D.