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| | + | [[009C Sample Midterm 2, Problem 5 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |-
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| − | |A geometric series <math>\sum_{n=0}^{\infty} ar^n</math> converges if <math style="vertical-align: -6px">|r|<1.</math>
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| − | |}
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| | + | [[009C Sample Midterm 2, Problem 5 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | '''Solution:'''
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| − | '''(a)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we notice that <math>\sum_{n=0}^\infty c_nx^n</math> is a geometric series.
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| − | |-
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| − | |We have <math style="vertical-align: -1px">r=x.</math>
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| − | |-
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| − | |Since this series converges,
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| − | |-
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| − | | <math>|r|=|x|<1.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |The series <math>\sum_{n=0} c_n\bigg(\frac{x}{2}\bigg)^n</math> is also a geometric series.
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| − | |-
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| − | |For this series, <math style="vertical-align: -13px">r=\frac{x}{2}.</math>
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| − | |-
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| − | |Now, we notice
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{|r|} & = & \displaystyle{\bigg|\frac{x}{2}\bigg|}\\
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| − | &&\\
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| − | & = & \displaystyle{\frac{|x|}{2}}\\
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| − | &&\\
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| − | & < & \displaystyle{\frac{1}{2}}
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| − | \end{array}</math>
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| − | |-
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| − | |since <math style="vertical-align: -5px">|x|<1.</math>
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| − | |-
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| − | | Since <math style="vertical-align: -5px">|r|<1,</math> this series converges.
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| − | |}
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| − |
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| − | '''(b)'''
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |-
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| − | |First, we notice that <math>\sum_{n=0}^\infty c_nx^n</math> is a geometric series.
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| − | |-
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| − | |We have <math style="vertical-align: -1px">r=x.</math>
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| − | |-
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| − | |Since this series converges,
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| − | |-
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| − | | <math>|r|=|x|<1.</math>
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |-
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| − | |The series <math>\sum_{n=0}^\infty c_n(-x)^n</math> is also a geometric series.
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| − | |-
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| − | |For this series, <math style="vertical-align: -1px">r=-x.</math>
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| − | |-
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| − | |Now, we notice
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| − | |-
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| − | <math>\begin{array}{rcl}
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| − | \displaystyle{|r|} & = & \displaystyle{|-x|}\\
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| − | &&\\
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| − | & = & \displaystyle{|x|}\\
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| − | &&\\
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| − | & < & \displaystyle{1}
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| − | \end{array}</math>
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| − | |-
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| − | |since <math style="vertical-align: -5px">|x|<1.</math>
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| − | |-
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| − | |Since <math style="vertical-align: -5px">|r|<1,</math> this series converges.
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| − | |}
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| − |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | | '''(a)''' converges (by the geometric series test)
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| − | |-
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| − | | '''(b)''' converges (by the geometric series test)
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| − | |}
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| | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
