SOLUTION: Can you solve this one? : ) 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? Answer:

Algebra ->  Sequences-and-series -> SOLUTION: Can you solve this one? : ) 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? Answer:       Log On


   

Question 87109: Can you solve this one? : )
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?
Answer:




b) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 10 terms? Carry all calculations to 6 decimals on all assignments.
Answer:


c) Using the formula for the sum of the first n terms of a geometric sequence, what is the sum of the first 12 terms? Carry all calculations to 6 decimals on all assignments.
Answer:




d) What observation can make about the successive partial sums of this sequence? In particular, what number does it appear that the sum will always be smaller than?
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 to find r, simply pick any term and divide it by the previous term:
r=%281%2F27%29%2F%281%2F9%29=%289%2F27%29=1%2F3 Divide the term of 1%2F27 by 1%2F9
So the ratio is
r=1%2F3
The sequence is reduced by a factor of 1%2F3 each term, so the sequence is %281%2F3%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%2F3%29%5E10%29%2F%281-%281%2F3%29%29 So plug in r=1%2F3 and n=10 to find the sum of the first 10 partial sums

S=%281-1%2F59049%29%2F%281-%281%2F3%29%29 Raise 1%2F3 to the 10 power

S=%2859049%2F59049-1%2F59049%29%2F%281%2F2%29 Make "1" into an equivalent fraction with a denominator of 59049

S=%2859048%2F59049%29%2F%281-%281%2F3%29%29 Subtract the fractions in the numerator

S=%2859048%2F59049%29%2F%282%2F3%29 Subtract the fractions in the denominator

S=177144%2F118098 Flip the 2nd fraction, multiply, and simplify

So the sum of the first ten terms is 177144%2F118098 or 1.49997459736829 approximately

note: I chose to use fractions (to maintain accuracy), but it may be much easier for you to simply use a calculator to evaluate the sum.
c)
The sum of a geometric series is

S=a%281-r%5En%29%2F%281-r%29where a=1

S=%281-%281%2F3%29%5E12%29%2F%281-%281%2F3%29%29 So plug in r=1%2F3 and n=12 to find the sum of the first 12 partial sums

S=%281-1%2F531441%29%2F%281-%281%2F3%29%29 Raise 1%2F3 to the 10 power

S=%28531441%2F59049-1%2F59049%29%2F%281%2F2%29 Make "1" into an equivalent fraction with a denominator of 531441

S=%28531440%2F531441%29%2F%281-%281%2F3%29%29 Subtract the fractions in the numerator

S=%28531440%2F531441%29%2F%282%2F3%29 Subtract the fractions in the denominator

S=1594320%2F1062882 Flip the 2nd fraction, multiply, and simplify

So the sum of the first twelve terms is 1594320%2F1062882 or 1.49999717748537 approximately

d)
It appears that the sums are approaching a finite number of 3%2F2 or 1.5. 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 sum of the infinite series. So if we let a=1 and r=1/3 we get
S=1%2F%281-%281%2F3%29%29
S=1%2F%282%2F3%29
S=3%2F2So this verifies that our series approaches 3%2F2.