SOLUTION: Im stumped!!! Is there an easy way to do these? Thanks so much! Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find the following: a) What is r, the ratio be

Algebra ->  Functions -> SOLUTION: Im stumped!!! Is there an easy way to do these? Thanks so much! Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find the following: a) What is r, the ratio be      Log On


   



Question 73789This question is from textbook college algebra
: Im stumped!!! Is there an easy way to do these? Thanks so much!



Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find the following:
a) What is r, the ratio between 2 consecutive terms?
Answer:
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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.
Answer:
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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.
Answer:
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d) What observation can make about these sums? In particular, what whole number does it appear that the sum will always be smaller than?
Answer:
This question is from textbook college algebra

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Starting with the sequence of: 1, 1/2, 1/4, 1/8,...
a) Find the common ratio.
The common ratio, r, is found by dividing any term by the preceeding term.
r+=+%281%2F2%29%2F1
r+=+1%2F2
or
r+=+%281%2F8%29%2F%281%2F4%29
r+=+%281%2F8%29%2A%284%2F1%29
r+=+1%2F2
The common ratio is 1/2
b) The partial sum of the first n terms of a geometric series is given by:
S%5Bn%5D+=+%28a%5B1%5D%281-r%5En%29%29%2F%281-r%29 where:
n is the number of the term (1st, 2nd, 3rd,...)
a%5B1%5D is the first term.
r is the common ratio.
To find the partial sum of the first 10 terms, set n+=+10, a%5B1%5D+=+1, r+=+1%2F2. Substitute these values into the formula for the partial sum.
S%5B10%5D+=+%281%281-%281%2F2%29%5E10%29%29%2F%281-%281%2F2%29%29
S%5B10%5D+=+0.9990%2F0.5
S%5B10%5D+=+1.9980 To four decimal places.
You should be able to finish the other parts using the above as a guide.
If you have trouble with it, please re-post.
c) Find the partial sum of the first 12 terms.
For this part, n = 12 and we can use the same formula for the partial sum of the first n terms of a geometric sequence.
S%5B12%5D+=+%281%281-%281%2F2%29%5E12%29%29%2F%281-1%2F2%29
S%5B12%5D+=+0.9998%2F%281%2F2%29
S%5B12%5D+=+1.9995 To four deicimal places.
d) My observation is that the sum approaches 2.
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