Nth term question?

Question

Find the sum of the first 16 terms of the geometric sequence 1/9,1/3,1, ....

Answer 1

The quick way to do this is to use the formula for the sum of the first n terms of a geometric series:

a = 1/9, r = 3
sum = a(1-r^n)/(1-r) = 1/9 * (1 - 3^16)/(1-3) = 2,391,484.444

The long way to do this is:

1/9 + 1/3 + 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 + 6561 + 19683 + 59049 + 177147 + 531441 + 1594323 = 2,391,484.444

Answer 2

S sub 16 = a sub 1(1-r^n)1-r r = 1/3 div by 1/9 = 3
S = 1/9[1-(3)^16]
S = 1/9 (1-43046721) / (1-3)


EDIT
This doesn't converge, it keeps increasing. The next term is 3, 9, 27, 81, 243, etc. The ratio is 3.

Answer 3

The sum of the first n terms of a geometric sequence can be given by:

[a(1)(1-r^n)]/(1-r)

In your case, a(1) is 1/9, and r=1/3 and n=16
The sum is, according to my TI83, .16666666628, or REALLY REALLY close to 1/6

--- EDIT ---
The above answers are not correct. This is a convergent sum.
It converges on 1/6

Answer 4

r=3 .The sixteenth term is 1/9*3^15
and the sum is
S = 1/9(3^16-1)/2 =2391484.4444