SOLUTION: 1. Use the geometric sequence of numbers 1, 1/3, 1/9, 1/27... to find the following: a. What is r, the ratio between 2 consecutive terms? (show work) b. Using the formula for the

1. Use the geometric sequence of numbers 1, 1/3, 1/9,
1/27... 
   to find the following:

a. What is r, the ratio between 2 consecutive terms? (show work)

2nd term ÷ 1st term = (1/3)÷1 = 1/3
3rd term ÷ 2nd term = (1/9)÷(1/3) = (1/9)×(3/1) = 3/9 = 1/3
4th term ÷ 3rd term = (1/27)÷(1/9) = (1/27)×(9/1) = 9/27 = 1/3

So the common ratio, r, must be 1/3

b. Using the formula for the sum of the first n terms of a 
   geometric series, 

      a1(1 - rn)
Sn = ------------
        1 - r

Where a1 is the first term, r the common ratio, and
n is the number of terms.

what is the sum of the first 10 terms?
Carry all calculations to 7 significant figures.
(show work)


       (1)[1 - (1/3)10]
S10 = ------------------
           1 - 1/3

       1 - 1/310
S10 = ----------- 
          2/3    

Punching that out on my TI-84

(1-1/3^10)/(2/3) ENTER

get .6666553766

or to 7 figures, .6666554

c. Using the formula for the sum of the first n terms of a
geometric series, what is the sum of the first 12 terms?

Carry all calculations to 7 significant figures. (show work)

       (1)[1 - (1/3)12]
S10 = ------------------
           1 - 1/3

       1 - 1/312
S10 = ----------- 
          2/3    

Punching that out on my TI-84

(1-1/3^12)/(2/3) ENTER

get .6666654122

or to 7 figures, .6666654

d. What observation can make about the successive partial 
sum of this series? In particular, what number does it 
appear that the sum will always be smaller than?

It appears that we get more and more 6's, and so it appears 
that the sum will always be smaller than .666666666666666... 
where the 6's go on forever. We know that this repeating 
decimal is equivalent to 2/3.  So it appears that the sum 
will always be less than 2/3.

Edwin