Difference between revisions of "008A Sample Final A, Question 14"

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(Created page with "'''Question: ''' Compute <math> \displaystyle{\sum_{n=1}^\infty 5\left(\frac{3}{5}\right)^n}</math> {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Founda...")
 
 
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Foundations
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!Foundations: &nbsp;
 
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|1) What type of series is this?
 
|1) What type of series is this?
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
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!Step 1: &nbsp;
 
|-
 
|-
 
|We start by identifying this series as a geometric series, and the desired formula for the sum being <math> S_\infty = \frac{a_1}{1 - r}</math>.  
 
|We start by identifying this series as a geometric series, and the desired formula for the sum being <math> S_\infty = \frac{a_1}{1 - r}</math>.  
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
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!Step 2: &nbsp;
 
|-
 
|-
 
|Since <math>a_1</math> is the first term in the series, <math>a_1 = 5\frac{3}{5} = 3</math>. The value for r is the ratio between consecutive terms, which is <math>\frac{3}{5}</math>. Plugging everything in we have <math> S_\infty = \frac{3}{1-\frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}</math>
 
|Since <math>a_1</math> is the first term in the series, <math>a_1 = 5\frac{3}{5} = 3</math>. The value for r is the ratio between consecutive terms, which is <math>\frac{3}{5}</math>. Plugging everything in we have <math> S_\infty = \frac{3}{1-\frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}</math>
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
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!Final Answer: &nbsp;
 
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|<math>\frac{15}{2}</math>
 
|<math>\frac{15}{2}</math>

Latest revision as of 23:01, 25 May 2015

Question: Compute

Foundations:  
1) What type of series is this?
2) Which formula, on the back page of the exam, is relevant to this question?
3) In the formula there are some placeholder variables. What is the value of each placeholder?
Answer:
1) This series is geometric. The giveaway is there is a number raised to the nth power.
2) The desired formula is
3) is the first term in the series, which is . The value for r is the ratio between consecutive terms, which is

Solution:

Step 1:  
We start by identifying this series as a geometric series, and the desired formula for the sum being .
Step 2:  
Since is the first term in the series, . The value for r is the ratio between consecutive terms, which is . Plugging everything in we have
Final Answer:  

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