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1. Find all \(r\) for which the infinite geometric series \(2 + 6r + 18r^2 + 54r^3 + \dotsb\)
is defined. Enter all possible values of \(r\) as an interval.

 

2. Compute \(1 - 2 + 3 - 4 + \dots + 2005 - 2006 + 2007\)

 

3. Let \(f(x) = \frac{x^2}{x^2 - 1}\) Find the largest integer \(n\) so that \(f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98\)

 Dec 6, 2019
 #1
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1. You want the geometric series to converge, so -1 < r < 1.

 

2. 1 - 2 + 3 - 4 + ... + 2005 - 2006 + 2007 = (1 - 2) + (3 - 4) + ... + (2005 - 2006) + 2007 = 1002.

 

3. The product will telescope, and you are left with n/(n - 1), so you want n/(n - 1) < 1.98.  The largest n that satisfies this is n = 201.

 Dec 6, 2019
 #2
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1)
Common ratio = \(\dfrac{6r}{2}\) = 3r

 

-1 < (Common ratio) < 1

 

-1 < 3r < 1

 

\(\dfrac{-1}{3} < r < \dfrac{1}{3}\)

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 Dec 6, 2019
 #3
avatar+7763 
+1

2)

Consider the pairwise sum of terms.

 

\(\quad 1-2+3-4+\cdots+2005-2006+2007\\ =(1-2)+(3-4)+\cdots+(2005-2006)+2007\\ =-1+(-1)+\cdots+(-1)+2007\\ \text{There are in total }\dfrac{2006}{2}=1003\text{ pairs that sums to }-1.\\ = -1(1003) + 2007\\ = 1004\)

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 Dec 6, 2019

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