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

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

Revision as of 22:33, 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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