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

2015-06-01T01:39:58Z

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...."

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''' 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;"
!Foundations
|-
|1) What type of series is this?
|-
|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?
|-
|Answer:
|-
|1) This series is geometric. The giveaway is there is a number raised to the nth power.
|-
|2) The desired formulas are <math>a_n = a\cdot r^{n-1}</math>   and   <math>S_\infty = \frac{a_1}{1-r}</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>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! Step 1:
|-
| The sequence is a geometric sequence. The common ratio is <math>r=\frac{-1}{3}</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! 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.
|-
| So, the formula for this geometric series is <math>a_n=(-3)\left(\frac{-1}{3}\right)^{n-1}</math>.
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
! 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;"
! Final Answer:
|-
| <math>a_n=(-3)\left(\frac{-1}{3}\right)^{n-1}</math>
|-
|<math>\frac{-9}{4}</math>
|}

[[005 Sample Final A|'''Return to Sample Exam''']]

005 Sample Final A, Question 22 - Revision history

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

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