Difference between revisions of "009C Sample Midterm 1, Problem 2"

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|Next, we calculate <math>s_3</math> and <math>s_4.</math> We have
 
|Next, we calculate <math>s_3</math> and <math>s_4.</math> We have
 
|-
 
|-
−
|
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 +
\displaystyle{s_3} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)}\\
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&&\\
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& = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^4}\bigg)}
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\end{array}</math>
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|-
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|and
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
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\displaystyle{s_4} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)+2\bigg(\frac{1}{2^4}-\frac{1}{2^5}\bigg)}\\
 +
&&\\
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& = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^5}\bigg).}
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\end{array}</math>
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|}
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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!Step 3: &nbsp;
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|-
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|If we look at <math>s_2,s_3,s_4, </math> we notice a pattern.
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|-
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|From this pattern, we get the formula
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|-
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|&nbsp; &nbsp; &nbsp; &nbsp; <math>s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).</math>
 
|}
 
|}
  

Revision as of 12:59, 12 February 2017

Consider the infinite series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=2}^\infty 2\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).}

a) Find an expression for the th partial sum of the series.
b) Compute Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\rightarrow \infty} s_n.}


Foundations:  
The Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} th partial sum, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n} for a series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^\infty a_n }
is defined as

        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n=\sum_{i=1}^n a_i.}

Solution:

(a)

Step 1:  
We need to find a pattern for the partial sums in order to find a formula.
We start by calculating Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_2} . We have
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_2=2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg).}
Step 2:  
Next, we calculate Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_3} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_4.} We have
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{s_3} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)}\\ &&\\ & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^4}\bigg)} \end{array}}
and
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{s_4} & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^3}\bigg)+2\bigg(\frac{1}{2^3}-\frac{1}{2^4}\bigg)+2\bigg(\frac{1}{2^4}-\frac{1}{2^5}\bigg)}\\ &&\\ & = & \displaystyle{2\bigg(\frac{1}{2^2}-\frac{1}{2^5}\bigg).} \end{array}}
Step 3:  
If we look at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_2,s_3,s_4, } we notice a pattern.
From this pattern, we get the formula
        Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n=2\bigg(\frac{1}{2^2}-\frac{1}{2^{n+1}}\bigg).}

(b)

Step 1:  
Step 2:  
Final Answer:  
(a)
(b)

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