Consider the infinite geometric series infinity sigma n=1 -4(1/3)^n-1. In

Consider the infinite geometric series below.a. Write the first 4 terms of the series b. Does the series diverge or converge? c.
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2. asked by lexxy
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Consider the infinite geometric series infinity sigma n=1 -4(1/3)^n-1. In this, the lower limit of the summation notion is
1. 1 answer
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1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4A)15 B)15/2 C)-15 D)1/15 I chose A 2)Find the sum of the infinite
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2. asked by Jon
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1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4A)15 B)15/2 C)-15 D)1/15 I chose A 2)Find the sum of the infinite
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2. asked by Jon
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how would I express the repeating decimal representaion of 1/9 as an infinite series using sigma notaion?Limit of (as n->
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2. asked by math!
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Determine whether the infinite series,sigma(((-1)^(n+1))/n)^2 converges or diverges. My professor gave these in a problem set
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2. asked by Ashley
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Consider the infinite geometric seriesn=1 infinity symbol -4(2/3)^n-1 a. Write the first four terms of the series. b. Does the
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Consider the infinite geometric seriesinfinity E -4(1/3)^n-1 n=1 In this image the lower limit of the summation notion is
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2. asked by grace
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Where does this infinite series converge:Sigma (k = 1 to infinity): 1/(9k^2 + 3k - 2)
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2. asked by Sean
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I'm having trouble with a geometric series problem.Determine if the infinite summation of (-3)^(n-1)/4^n converges or diverges.
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2. asked by Jace
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