005 Sample Final A, Question 22 - Revision history

Matt Lee at 04:23, 22 May 2015

2015-05-22T04:23:52Z

Matt Lee at 04:16, 22 May 2015

2015-05-22T04:16:36Z

Kayla Murray at 20:03, 20 May 2015

2015-05-20T20:03:06Z

Kayla Murray at 19:20, 20 May 2015

2015-05-20T19:20:48Z

Matt Lee: Created page with "''' Question ''' Consider the following sequence,
$-3, 1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \cdots$
a...."

2015-05-01T04:11:02Z

Created page with "''' Question ''' Consider the following sequence, <center><math> -3, 1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \cdots </math></center> a...."

New page

''' Question ''' Consider the following sequence, 
<center><math> -3, 1, -\frac{1}{3}, \frac{1}{9}, -\frac{1}{27}, \cdots </math></center> 
     a. Determine a formula for <math>a_n</math>, the n-th term of the sequence. 
     b. Find the sum <math> \displaystyle{\sum_{k=1}^\infty a_k}</math>

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Final Answers
|-
|a) False. Nothing in the definition of a geometric sequence requires the common ratio to be always positive. For example, <math>a_n = (-a)^n</math>
|-
|b) False. Linear systems only have a solution if the lines intersect. So y = x and y = x + 1 will never intersect because they are parallel.
|-
|c) False. <math>y = x^2</math> does not have an inverse.
|-
|d) True. <math>cos^2(x) - cos(x) = 0</math> has multiple solutions.
|-
|e) True.
|-
|f) False.
|}

@@ Line 11: / Line 11: @@
 |1) What type of series is this?
 |-
-|2) Which formula, on the back page of the exam, is relevant to this question?
+|2) Which formulas, about this type of series, are relevant to this question?
 |-
 |3) In the formula there are some placeholder variables. What is the value of each placeholder?
@@ Line 19: / Line 19: @@
 |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>
+|2) The desired formulas are <math>a_n = a\cdot r^{n-1}</math> &nbsp; and &nbsp; <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>
+|3) <math>a_1</math> is the first term in the series, which is <math> -3</math>. The value for r is the ratio between consecutive terms, which is <math>\frac{-1}{3}</math>
 |}

@@ Line 3: / Line 3: @@
 &nbsp;&nbsp;&nbsp;&nbsp; a. Determine a formula for <math>a_n</math>, the n-th term of the sequence. <br>
 &nbsp;&nbsp;&nbsp;&nbsp; b. Find the sum <math> \displaystyle{\sum_{k=1}^\infty a_k}</math>
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"

@@ Line 7: / Line 7: @@
 ! Step 1:
 |-
+| The sequence is a geometric sequence. The common ratio is <math>r=\frac{-1}{3}</math>.
 |}
@@ Line 13: / Line 13: @@
 ! Step 2:
 |-
+| The formula for the nth term of a geometric series is <math>a_n=ar^{n-1}</math> where <math>a</math> is the first term of the sequence.
 |}
@@ Line 19: / Line 21: @@
 ! Step 3:
 |-
+| For geometric series, <math>\displaystyle{\sum_{k=1}^\infty a_k}=\frac{a}{1-r}</math> if <math>|r|<1</math>. Since <math>|r|=\frac{1}{3}</math>,
 |-
+| we have <math>\displaystyle{\sum_{k=1}^\infty a_k}=\frac{-3}{1-\frac{-1}{3}}=\frac{-9}{4}</math>.
 |}
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
-! Step 4:
+! Final Answer:
 |-
+| <math>a_n=(-3)\left(\frac{-1}{3}\right)^{n-1}</math>
 |-
+|<math>\frac{-9}{4}</math>
 |}