Difference between revisions of "009C Sample Final 2, Problem 2"

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!Step 1:    
 
!Step 1:    
 
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|-
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|Let &nbsp;<math>a_n</math>&nbsp; be the &nbsp;<math>n</math>th term of this sum.
 +
|-
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|We notice that
 +
|-
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|&nbsp; &nbsp; &nbsp, &nbsp; <math>\frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},</math>&nbsp; and &nbsp;<math>\frac{a_4}{a_2}=\frac{-1}{2}.</math>
 
|-
 
|-
|
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|So, this is a geometric series with &nbsp;<math>r=\frac{-1}{2}.</math>
 
|-
 
|-
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|Since &nbsp;<math>|r|<1,</math>&nbsp; this series converges.
 
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|}
  
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
|
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|Hence, the sum of this geometric series is
 
|-
 
|-
 
|
 
|
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&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{4}{1-(-\frac{1}{2})}}\\
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&&\\
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& = & \displaystyle{\frac{4}{\frac{3}{2}}}\\
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&&\\
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& = & \displaystyle{\frac{8}{3}.}
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\end{array}</math>
 
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|}
  
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
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|&nbsp;&nbsp; '''(a)'''  
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|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp;<math>\frac{8}{3}</math>
 
|-
 
|-
 
|&nbsp;&nbsp; '''(b)'''  
 
|&nbsp;&nbsp; '''(b)'''  
 
|}
 
|}
 
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]
 
[[009C_Sample_Final_2|'''<u>Return to Sample Exam</u>''']]

Revision as of 18:49, 4 March 2017

For each of the following series, find the sum if it converges. If it diverges, explain why.

(a)  

(b)  

Foundations:  
1. The sum of a convergent geometric series is   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a}{1-r}}
        where  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r}   is the ratio of the geometric series
        and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a}   is the first term of the series.
2. The  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th partial sum,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n}   for a series  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty a_n }   is defined as

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n=\sum_{i=1}^n a_i.}


Solution:

(a)

Step 1:  
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n}   be the  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th term of this sum.
We notice that
    &nbsp,   Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a_2}{a_1}=\frac{-2}{4},~\frac{a_3}{a_2}=\frac{1}{-2},}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{a_4}{a_2}=\frac{-1}{2}.}
So, this is a geometric series with  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=\frac{-1}{2}.}
Since  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|<1,}   this series converges.
Step 2:  
Hence, the sum of this geometric series is

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{4}{1-(-\frac{1}{2})}}\\ &&\\ & = & \displaystyle{\frac{4}{\frac{3}{2}}}\\ &&\\ & = & \displaystyle{\frac{8}{3}.} \end{array}}

(b)

Step 1:  
Step 2:  


Final Answer:  
   (a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{8}{3}}
   (b)

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