SOLUTION: 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: r=1/2 divided by 1, which

Algebra ->  Sequences-and-series -> SOLUTION: 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: r=1/2 divided by 1, which       Log On


   

Question 84140: 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: r=1/2 divided by 1,
which is 1/2(fraction)
Is this correct
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? Round your answer to 4 decimals.
show work
1st term is 1X1/2=1/2
2nd term is 1/2X1/2=1/4
3rd term is 1/4X1/2=1/8
4th term is 1/8X1/2=1/16
5th term is 1/16X1/2=1/32
6th term is 1/32X1/2=1/64
7th term is 1/64X1/2=1/128
8th term is 1/128X1/2=1/256
9th term is 1/256X1/2=1/512
10th term is 1/512X1/2=1,024
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? Round to 4 decimals.
show work
11th term is 1/1,024X1/2=2,048
12 term is 1/2,048X1/2=4,096
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?
1
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
So the ratio is r=1%2F2 (you are correct)
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.