|
|
| (9 intermediate revisions by the same user not shown) |
| Line 1: |
Line 1: |
| | <span class="exam">Evaluate: | | <span class="exam">Evaluate: |
| | | | |
| − | ::<span class="exam">a) <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math>
| + | <span class="exam">(a) <math>\lim _{n\rightarrow \infty} \frac{1}{\big(\frac{n-4}{n}\big)^n}</math> |
| − | ::<span class="exam">b) <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math>
| |
| | | | |
| | + | <span class="exam">(b) <math>\sum_{n=1}^\infty \frac{1}{2} \bigg(\frac{1}{4}\bigg)^{n-1} </math> |
| | + | <hr> |
| | + | [[009C Sample Midterm 2, Problem 1 Solution|'''<u>Solution</u>''']] |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Foundations:
| |
| − | |-
| |
| − | |
| |
| − | |-
| |
| − | |Sum formula for geometric series
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| | | | |
| − | '''Solution:''' | + | [[009C Sample Midterm 2, Problem 1 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | '''(a)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |
| |
| − | |-
| |
| − | |
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |First, we not that this is a geometric series with <math>r=\frac{1}{4}.</math>
| |
| − | |-
| |
| − | |Since <math>|r|=\frac{1}{4}<1,</math>
| |
| − | |-
| |
| − | |this series converges.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |Now, we need to find the sum of this series.
| |
| − | |-
| |
| − | |The first term of the series is <math>a_1=\frac{1}{2}.</math>
| |
| − | |-
| |
| − | |Hence, the sum of the series is
| |
| − | |-
| |
| − | |
| |
| − | <math>\begin{array}{rcl}
| |
| − | \displaystyle{\frac{a_1}{1-r}} & = & \displaystyle{\frac{\frac{1}{2}}{1-\frac{1}{4}}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\frac{\big(\frac{1}{2}\big)}{\big(\frac{3}{4}\big)}}\\
| |
| − | &&\\
| |
| − | & = & \displaystyle{\frac{2}{3}}
| |
| − | \end{array}</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | |'''(a)'''
| |
| − | |-
| |
| − | | '''(b)''' <math>\frac{2}{3}</math>
| |
| − | |}
| |
| | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |
