SOLUTION: Find the sum of the infinite geometric series 3/5-1/5+1/15-1/45+... 1+1/4+1/16+... 64+16+4+... 1/4+1/16+1/64+... The following problem refers to a geometric sequenc

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Find the sum of the infinite geometric series 3/5-1/5+1/15-1/45+... 1+1/4+1/16+... 64+16+4+... 1/4+1/16+1/64+... The following problem refers to a geometric sequenc      Log On


   

Question 1160668: Find the sum of the infinite geometric series
3/5-1/5+1/15-1/45+...
1+1/4+1/16+...
64+16+4+...
1/4+1/16+1/64+...
The following problem refers to a geometric sequence:
If a_1= -1 and r= -1, find a_45
If a_1= 3 and r= 5, find a_n
Determine if the following sequence is geometric progressions. If it is, identify the common ratio r, if
not answer DNE.
1/5,-1/15, 1/45, -1/135,...

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

I'll do the first two problems to get you started.

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Problem 1

The first term is a = 3/5
The common ratio is r = -1/3

We multiply each term by -1/3 to get the next term
first term = -3/5
second term = (common ratio)*(first term) = (-1/3)*(3/5) = -1/5
third term = (common ratio)*(second term) = (-1/3)*(-1/5) = 1/15
etc etc

With r = -1/3 = -0.33 (approximately), we can see that -1 < r < 1 is true, which means that the infinite geometric sum converges to a finite number. That number is...
S+=+a%2F%281-r%29

S+=+%283%2F5%29%2F%281-%28-1%2F3%29%29

S+=+%283%2F5%29%2F%281%2B1%2F3%29

S+=+%283%2F5%29%2F%283%2F3%2B1%2F3%29

S+=+%283%2F5%29%2F%284%2F3%29

S+=+%283%2F5%29%2A%283%2F4%29

S+=+9%2F20


Answer: 9/20

With your calculator, note how
  • 3/5-1/5 = 0.4
  • 3/5-1/5+1/15 = 0.46666666666667 (approximate)
  • 3/5-1/5+1/15-1/45 = 0.44444444444444 (approximate)
  • 3/5-1/5+1/15-1/45+1/135 = 0.45185185185186 (approximate)
  • 3/5-1/5+1/15-1/45+1/135-1/405 = 0.44938271604939 (approximate)
  • 3/5-1/5+1/15-1/45+1/135-1/405+1/1215 = 0.45020576131688 (approximate)
Those partial sums are slowly approaching 9/20 = 0.45; they'll never actually get there because we can't actually reach infinity. The sums only get closer and closer.

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Problem 2

Same idea as problem 1. We have different values this time of course. In this problem,
a = 1 is the first term
r = 1/4 is the common ratio

We can determine the common ratio by picking any term but the first one, then dividing it over its previous term

examples:
(second term)/(first term) = (1/4) divided by (1) = 1/4
(third term)/(second term) = (1/16) divided by (1/4) = (1/16)*(4/1) = 4/16 = 1/4
These are two of many ways to see why the common ratio is r = 1/4.

The infinite sum converges to a finite value because r = 1/4 = 0.25 makes -1 < r < 1 true.

Side note: We can condense -1 < r < 1 into the absolute value inequality |r| < 1 . This says r is within 1 unit of zero, or that the distance from r to 0 is less than 1.

Anyways, back to the problem. We plug a = 1 and r = 1/4 into the formula used in the prior question.

S+=+a%2F%281-r%29

S+=+%281%29%2F%281-1%2F4%29

S+=+1%2F%284%2F4-1%2F4%29

S+=+1%2F%283%2F4%29

S+=+1%2A%284%2F3%29

S+=+4%2F3

Answer: 4/3

Partial check or partial confirmation we have the right answer:
  • 1+1/4 = 1.25
  • 1+1/4+1/16 = 1.3125
  • 1+1/4+1/16+1/64 = 1.328125
  • 1+1/4+1/16+1/64+1/256 = 1.33203125
  • 1+1/4+1/16+1/64+1/256+1/1024 = 1.3330078125
  • 1+1/4+1/16+1/64+1/256+1/1024+1/4096 = 1.333251953125
The partial sums are gradually approaching 4/3 = 1.33333...