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Calcular ingeniosamente el siguiente producto (1+2/3)*(1+2/4)*(1+2/5)*(1+2/6)*(1+2/7)*(1+2/8)*(1+2/9)*(1+2/10)*(1+2/11)*(1+2/12)*(1+2/13)*(1+2/14)*(1+2/15)(1+2/16)*(1+2/17)(1+2/18)*(1+2/19)*(1+2/20)(1+2/21)*(1+2/22)

 28 jun 2014

Mejor Respuesta  

 #1
avatar+118587 
+8

=$$\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{4}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{7}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{6}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{7}}}}\right)$$   etc up to $$\left({\frac{{\mathtt{23}}}{{\mathtt{21}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{24}}}{{\mathtt{22}}}}\right)$$

 

now the fives cancel, the sixes cancel, the 7 cancels, etc

Ahora los cincos cancelar, los seises cancelar 7 cancela, etc.

 

$${\frac{\left({\mathtt{23}}{\mathtt{\,\times\,}}{\mathtt{24}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{4}}\right)}}$$  

=  46

 

 
 28 jun 2014
 #1
avatar+118587 
+8
Mejor Respuesta

=$$\left({\frac{{\mathtt{5}}}{{\mathtt{3}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{6}}}{{\mathtt{4}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{7}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{8}}}{{\mathtt{6}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{9}}}{{\mathtt{7}}}}\right)$$   etc up to $$\left({\frac{{\mathtt{23}}}{{\mathtt{21}}}}\right){\mathtt{\,\times\,}}\left({\frac{{\mathtt{24}}}{{\mathtt{22}}}}\right)$$

 

now the fives cancel, the sixes cancel, the 7 cancels, etc

Ahora los cincos cancelar, los seises cancelar 7 cancela, etc.

 

$${\frac{\left({\mathtt{23}}{\mathtt{\,\times\,}}{\mathtt{24}}\right)}{\left({\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{4}}\right)}}$$  

=  46

 

 
Melody 28 jun 2014

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