SOLUTION: Can someone assist me with the following problem please? 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 consecut

Algebra ->  Sequences-and-series -> SOLUTION: Can someone assist me with the following problem please? 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 consecut      Log On


   

Question 83789: Can someone assist me with the following problem please?
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?
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
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.
d) What observation can make about these sums? In particular, what whole number does it appear that the sum will always be smaller than?
Thank you for your assistance...
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)
The ratio r is the factor to get from term to term. So
r=nth term/(n-1) term
r=%281%2F8%29%2F%281%2F4%29=%284%2F8%29=1%2F2
r=1%2F2
The sequence is cut in half each term, so the sequence is %281%2F2%29%5En

b)
The sum of a geometric series is
S=a%281-r%5En%29%2F%281-r%29where a=1
S=%281-%281%2F2%29%5E10%29%2F%281-%281%2F2%29%29So plug in n=10 to find the sum of the first 10 partial sums
S=%281-1%2F1024%29%2F%281%2F2%29
S=2046%2F1024
So the sum of the first ten terms is 2046%2F1024 or 1.99805 approximately
c)
Use the same formula to find the sum of the 1st 12 terms
S=a%281-r%5En%29%2F%281-r%29where a=1
S=%281-%281%2F2%29%5E12%29%2F%281-%281%2F2%29%29So plug in n=12 to find the sum of the first 12 partial sums
S=%281-1%2F4096%29%2F%281%2F2%29
S=8190%2F4096
So the sum of the first twelve terms is 8190%2F4096 or 1.99951 approximately

d)
It appears that the sums are approaching a finite number of 2. This is because each term is getting smaller and smaller. This observation is justified by the fact that if abs%28r%29%3C1 then the infinite series will approach a finite number. In other words
If abs%28r%29%3C1 (the magnitude of r has to be less than 1) then,
S=a%2F%281-r%29Where S is the infinite series. So if we let a=1 and r=1/2 we get
S=1%2F%281-%281%2F2%29%29
S=1%2F%281%2F2%29
S=2So this verifies that our series approaches 2. Hope this helps.