009C Sample Final 1, Problem 2 - Revision history

Kayla Murray at 00:57, 3 December 2017

2017-12-03T00:57:39Z

Kayla Murray at 22:57, 26 February 2017

2017-02-26T22:57:08Z

Kayla Murray at 22:50, 26 February 2017

2017-02-26T22:50:50Z

Kayla Murray at 23:51, 25 February 2017

2017-02-25T23:51:58Z

Kayla Murray at 23:50, 25 February 2017

2017-02-25T23:50:23Z

Kayla Murray at 23:46, 25 February 2017

2017-02-25T23:46:21Z

Kayla Murray at 23:44, 25 February 2017

2017-02-25T23:44:52Z

Kayla Murray at 01:26, 19 April 2016

2016-04-19T01:26:33Z

Kayla Murray at 18:28, 1 March 2016

2016-03-01T18:28:42Z

Kayla Murray at 18:48, 29 February 2016

2016-02-29T18:48:55Z

@@ Line 5: / Line 5: @@
 <span class="exam">(b) &nbsp; <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
+<hr>
-!Foundations: &nbsp;
+[[009C Sample Final 1, Problem 2 Solution|'''<u>Solution</u>''']]
-'''Solution:'''
+[[009C Sample Final 1, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']]
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 9: / Line 9: @@
 |-
+|
-'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math>
+'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> &nbsp; with &nbsp; <math>|r|<1,</math>
 |-
+|
@@ Line 15: / Line 15: @@
 |-
+|
-'''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math>
+'''2.''' For a telescoping series, we find the sum by first looking at the partial sum &nbsp; <math style="vertical-align: -3px">s_k</math>
 |-
+|
@@ Line 42: / Line 42: @@
 !Step 2: &nbsp;
 |-
-|Since <math style="vertical-align: -16px">2<e,~\bigg|-\frac{2}{e}\bigg|<1.</math> So,
+|Since &nbsp; <math style="vertical-align: -16px">2<e,~\bigg|-\frac{2}{e}\bigg|<1.</math>
 |-
+|
@@ Line 61: / Line 63: @@
 |This is a telescoping series. First, we find the partial sum of this series.
 |-
-|Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
+|Let &nbsp; <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
 |-
 |Then,

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the sum of the following series:
-<span class="exam">(a) <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math>
+<span class="exam">(a) &nbsp; <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math>
-<span class="exam">(b) <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>
+<span class="exam">(b) &nbsp; <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 32: / Line 32: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\sum_{n=0}^{\infty} (-2)^n e^{-n}} & = & \displaystyle{\sum_{n=0}^{\infty} \frac{(-2)^n}{e^n}}\\
 &&\\
@@ Line 45: / Line 45: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\sum_{n=0}^{\infty} (-2)^ne^{-n}} & = & \displaystyle{\frac{1}{1+\frac{2}{e}}}\\
 &&\\
@@ Line 66: / Line 66: @@
 |-
+|
-::<math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.</math>
 |}
@@ Line 75: / Line 75: @@
 |-
+|
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 \displaystyle{\sum_{n=1}^{\infty}\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)} & = & \displaystyle{\lim_{k\rightarrow \infty} s_k}\\
 &&\\

@@ Line 7: / Line 7: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-
+|
-::'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math>
+'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math>
 |-
+|
-:::<math>\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;<math>\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.</math>
 |-
+|
-::'''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math>
+'''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math>
 |-
+|
-:::and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
+&nbsp; &nbsp; &nbsp; &nbsp;and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
 |}

@@ Line 22: / Line 22: @@
 :::and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
 |}
 '''Solution:'''
@@ Line 84: / Line 85: @@
 \end{array}</math>
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|&nbsp;&nbsp; '''(a)''' <math>\frac{e}{e+2}</math>
+|&nbsp;&nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{e}{e+2}</math>
 |-
-|&nbsp;&nbsp; '''(b)''' <math>\frac{1}{2}</math>
+|&nbsp;&nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{1}{2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

@@ Line 1: / Line 1: @@
 <span class="exam"> Find the sum of the following series:
-<span class="exam">a) <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math>
+::<span class="exam">a) <math>\sum_{n=0}^{\infty} (-2)^ne^{-n}</math>
-<span class="exam">b) <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>
+::<span class="exam">b) <math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
@@ Line 10: / Line 10: @@
 |Recall:
 |-
 |'''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math>
 |-
+|
-::<math>\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.</math>
+:::<math>\sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.</math>
 |-
 |'''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math>
 |-
+|
-::and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
+:::and then calculate <math style="vertical-align: -14px">\lim_{k\rightarrow\infty} s_k.</math>
 |}
@@ Line 44: / Line 46: @@
 |-
+|
-::<math>\sum_{n=0}^{\infty} (-2)^ne^{-n}=\frac{1}{1+\frac{2}{e}}=\frac{1}{\frac{e+2}{e}}=\frac{e}{e+2}.</math>
+::<math>\begin{array}{rcl}
 |}
@@ Line 56: / Line 64: @@
 |Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math>
 |-
-|Then, <math style="vertical-align: -14px">s_k=\frac{1}{2}-\frac{1}{2^{k+1}}.</math>
+|Then,
 |}
@@ Line 65: / Line 76: @@
 |-
+|
-::<math>\sum_{n=1}^{\infty} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)=\lim_{k\rightarrow \infty} s_k=\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}=\frac{1}{2}.</math>
+::<math>\begin{array}{rcl}
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Final Answer: &nbsp;
 |-
-|'''(a)''' <math>\frac{e}{e+2}</math>
+|&nbsp;&nbsp; '''(a)''' <math>\frac{e}{e+2}</math>
 |-
-|'''(b)''' <math>\frac{1}{2}</math>
+|&nbsp;&nbsp; '''(b)''' <math>\frac{1}{2}</math>
 |}
 [[009C_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]