8A F11 Q14 - Revision history

Matt Lee at 15:26, 8 April 2015

2015-04-08T15:26:34Z

Matt Lee at 23:00, 6 April 2015

2015-04-06T23:00:31Z

Matt Lee: Created page with "'''Question: ''' Compute $\displaystyle{\sum_{n=1}^\infty 5\left(\frac{3}{5}\right)^n}$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Founda..."

2015-03-29T17:40:00Z

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..."

New page

'''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;"
!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 <math>S_\infty = \frac{a_1}{1-r}</math>
|-
|3) <math>a_1</math> is the first term in the series, which is <math> 5\frac{3}{5} = 3</math>. The value for r is the ratio between consecutive terms, which is <math>\frac{3}{5}</math>
|}

Solution:

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
|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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 2:
|-
|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>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answer:
|-
|Plugging everything in we have <math> S_\infty = \frac{3}{1-\frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}</math>
|}

← Older revision		Revision as of 23:00, 6 April 2015
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	\|Plugging everything in we have <math> S_\infty = \frac{3}{1-\frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}</math>		\|Plugging everything in we have <math> S_\infty = \frac{3}{1-\frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}</math>
	\|}		\|}
		+
		+	[[8AF11Final\|<u>'''Return to Sample Exam</u>''']]

@@ Line 30: / Line 30: @@
 ! Step 2:
 |-
-|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>.
+|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>
 |}
@@ Line 36: / Line 36: @@
 ! Final Answer:
 |-
-|Plugging everything in we have <math> S_\infty = \frac{3}{1-\frac{3}{5}} = \frac{3}{\frac{2}{5}} = \frac{15}{2}</math>
+|<math>\frac{15}{2}</math>
 |}
 [[8AF11Final|<u>'''Return to Sample Exam</u>''']]

8A F11 Q14 - Revision history

Matt Lee at 15:26, 8 April 2015

Matt Lee at 23:00, 6 April 2015

Matt Lee: Created page with "'''Question: ''' Compute \displaystyle{\sum_{n=1}^\infty 5\left(\frac{3}{5}\right)^n} {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Founda..."

Matt Lee: Created page with "'''Question: ''' Compute $\displaystyle{\sum_{n=1}^\infty 5\left(\frac{3}{5}\right)^n}$ {| class="mw-collapsible mw-collapsed" style = "text-align:left;" !Founda..."