Question 64422
<pre><font size = 5><b>
Use the geometric sequence of numbers 
1, 1/2, 1/4, 1/8,...to 
find the following:
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a) What is r, the ratio between 2 consecutive terms? 


Just divide each given term  after the first by the 
preceding one and see if you get the same number.  If 
you do, then you call that number "the common ratio, r".

For 1, 1/2, 1/4, 1/8,... we divide the second term, 1/2,
by the first term 1, like this: 1/2÷1 = 1/2.  Then we 
divide the third term 1/4, by the second term 1/2, like 
this: (1/4)÷(1/2) = (1/4)×(2/1) = 2/4 = 1/2. 
Then we divide the fourth term, 1/8, by the third term, 
1/4, like this" (1/8)÷(1/4) = (1/8)×(4/1) = 4/8 = 1/2.  

Every time we got 1/2. So that means this is a geometric 
sequence and the common ratio, r, is 1/2.  So r = 1/2. 

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b) Using the formula for the sum of the first n 
terms of a geometric series, what is the sum of 
the first 10 terms? Please round your answer to 
4 decimals.

The formula for the sum, called S<sub>n</sub>, of the 
first n terms of a geometric sequence is either 
of these two equivalent formulas:

S<sub>n</sub> = a<sub>1</sub>(r<sup>n</sup> - 1)/(r - 1)

or

S<sub>n</sub> = a<sub>1</sub>(1 - r<sup>n</sup>)/(1 - r)

where a<sub>1</sub> stands for the first term, r stands 
for the common ratio, and n stands for the number of 
term that you want to find.  It doesn't matter which of 
those formulas you use, because you'll get the same 
answer using either one. Normally we use the first one 
if |r| > 1 and the second one if |r| < 1, but there is 
no rule. I'll use the second one. 
 
Here a<sub>1</sub> = 1, r = 1/2, and n = 10 so we plug those in:

S<sub>n</sub> = a<sub>1</sub>(1 - r<sup>n</sup>)/(1 - r)

S<sub>10</sub> = (1)[1 - (1/2)<sup>10</sup>]/(1 - 1/2)

S<sub>10</sub> = (1)(1 - 1/2<sup>10</sup>)/(1/2)

S<sub>10</sub> = (1 - 1/2<sup>10</sup>)/(1/2)

S<sub>10</sub> = (1 - 1/2<sup>10</sup>)×(2/1)

S<sub>10</sub> = 2(1 - 1/2<sup>10</sup>)

S<sub>10</sub> = 2 - 2/2<sup>10</sup>

S<sub>10</sub> = 2 - 1/2<sup>9</sup> 

S<sub>10</sub> = 2 - 1/512 = 1.9980 

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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? Please round your answer to 
4 decimals.

Same as above using 12 for n instead of 10

a<sub>1</sub> = 1, r = 1/2, and n = 12 so we plug those in:

S<sub>n</sub> = a<sub>1</sub>(1 - r<sup>n</sup>)/(1 - r)

S<sub>12</sub> = (1)[1 - (1/2)<sup>12</sup>]/(1 - 1/2)

S<sub>12</sub> = (1)(1 - 1/2<sup>12</sup>)/(1/2)

S<sub>12</sub> = (1 - 1/2<sup>12</sup>)/(1/2)

S<sub>12</sub> = (1 - 1/2<sup>12</sup>)×(2/1)

S<sub>12</sub> = 2(1 - 1/2<sup>12</sup>)

S<sub>12</sub> = 2 - 2/2<sup>12</sup>

S<sub>12</2ub> = 2 - 1/2<sup>11</sup> 

S<sub>12</sub> = 2 - 1/2048 = 1.9995 

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<pre><font size = 6>
d) What observation can you make about these 
sums? In particular, what whole number does 
it appear that the sum will always be smaller 
than?

Well, we found:

S<sub>10</sub> = 2 - 1/512 = 1.9980 

S<sub>12</sub> = 2 - 1/2048 = 1.9995 

It appears that each time we will be
subtracting a smaller fraction away
from 2, so the whole number that the
it appears the sum will always be a
little smaller than -- is 2.

Edwin</pre></font></b>