Difference between revisions of "2013 AMC 12A Problems/Problem 4"

(Created page with "We can factor a 2^2012 out of the numerator and denominator to obtain ((2^2012)((2^2)+1))/((2^2012)((2^2)-1)) Both the (2^2012) terms cancel, so we get ((2^2)+1)/((2^2)-1) = ...")
 
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<math>\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}</math>
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We can factor a 2^2012 out of the numerator and denominator to obtain
 
We can factor a 2^2012 out of the numerator and denominator to obtain
  
((2^2012)((2^2)+1))/((2^2012)((2^2)-1))
+
<math>\frac{2^{2012}*(2^2+1)}{2^{2012}*(2^2-1)}</math>
  
 
Both the (2^2012) terms cancel, so we get  
 
Both the (2^2012) terms cancel, so we get  
  
((2^2)+1)/((2^2)-1) = 5/3, which is C
+
<math>\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}</math>, which is C

Revision as of 22:17, 6 February 2013

$\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}$

We can factor a 2^2012 out of the numerator and denominator to obtain

$\frac{2^{2012}*(2^2+1)}{2^{2012}*(2^2-1)}$

Both the (2^2012) terms cancel, so we get

$\frac{(2^2+1)}{(2^2-1)}=\frac{5}{3}$, which is C

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