Question 310633: I can't remember how to solve the infinite series.
1. What is the solution to the infinite geometric series... the lower limit is = to 1. The upper limit is the infinite symbol. 0.8(0.2)^(n-1).
Now I know the formula is a1/1 - r. But I can't remember which goes for A1 and which goes for r.
2. What is the solution to the infinite geometric series... the lower limit is = to 1. The upper limit is the infinite symbol. (-2/3)^(n-1)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. What is the solution to the infinite geometric series... the lower limit is = to 1. The upper limit is the infinite symbol. 0.8(0.2)^(n-1).
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S = a(1)[1/(1-r)
a(1) = 0.8(0.2)^(0) == 0.8
r = 0.2
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S = 0.8[1-0.2] = 1
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2. What is the solution to the infinite geometric series... the lower limit is = to 1. The upper limit is the infinite symbol. (-2/3)^(n-1)
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This series alternates : 1, -2/3, 4/9, -8/27, 16/81... etc.
So you have two series:
1, 4/9, 16/81,...
and
(-2/3), (-8/27)...
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The sum of the 1st = ?
a(1) = 1
r = 4/9
S = 1/(1- 4/9) = 1/(5/9) = 9/5
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The sum of the 2nd = ?
a(1) = -2/3r = (4/9)
S = (-2/3)/[1-4/9] = (-2/3)/(5/9) = -6/5
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Total sum = (9/5)-(6/5) = 3/5
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Note: You can find reminders of how to handle virtually
all math procedures by using Google on the web.
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Cheers,
Stan H.
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