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avatar+60 

\(\frac{1}{2*4}+\frac{1}{6*4}+\frac{1}{6*8}+\frac{1}{10*8}+\cdots+\frac{1}{96*98}+\frac{1}{100*98}\)

\(\frac{1}{11*15}+\frac{1}{19*15}+\frac{1}{19*23}+\frac{1}{23*27}+\cdots+\frac{1}{1103·1107}+\frac{1}{1111·1107}\)

 

 

I have no idea how to solve these. If you could help, that would be great!!

 Dec 10, 2018
 #1
avatar+21855 
+11

Calculate:

\(\large{\frac{1}{2*4}+\frac{1}{6*4}+\frac{1}{6*8}+\frac{1}{10*8}+\cdots+\frac{1}{96*98}+\frac{1}{100*98}}\)

 

\(\begin{array}{|rcll|} \hline && \frac{1}{2*4}+\frac{1}{4*6}+\frac{1}{6*8}+\frac{1}{8*10}+\frac{1}{10*12}+\frac{1}{12*14}+\cdots+\frac{1}{94*96}+\frac{1}{96*98}+\frac{1}{98*100} \\\\ &=& \frac12\left( \frac{1}{2}-\frac{1}{4}\right) +\frac12\left( \frac{1}{4}-\frac{1}{6}\right) +\frac12\left( \frac{1}{6}-\frac{1}{8}\right) +\frac12\left( \frac{1}{8}-\frac{1}{10}\right) \\ &&+\frac12\left( \frac{1}{10}-\frac{1}{12}\right) +\frac12\left( \frac{1}{12}-\frac{1}{14}\right)+\cdots\\ && +\frac12\left( \frac{1}{94}-\frac{1}{96}\right) +\frac12\left( \frac{1}{96}-\frac{1}{98}\right) +\frac12\left( \frac{1}{98} -\frac{1}{100} \right)\\\\ &=& \frac14 \\ &&-\frac12\left(\frac{1}{4}\right)+ \frac12\left(\frac{1}{4}\right)\\ &&-\frac12\left(\frac{1}{6}\right)+\frac12\left(\frac{1}{6}\right) \\ &&-\frac12\left(\frac{1}{8}\right)+\frac12\left(\frac{1}{8}\right) \\ &&-\frac12\left( \frac{1}{10}\right)+\frac12\left( \frac{1}{10}\right) \\ &&-\frac12\left( \frac{1}{12}\right)+\frac12\left( \frac{1}{12}\right) \\ &&-\frac12\left( \frac{1}{14}\right)+\frac12\left( \frac{1}{14}\right)+\cdots \\ &&-\frac12\left( \frac{1}{94}\right)+\frac12\left( \frac{1}{94}\right)\\ &&-\frac12\left( \frac{1}{96}\right)+\frac12\left( \frac{1}{96}\right) \\ &&-\frac12\left( \frac{1}{98}\right)+\frac12\left( \frac{1}{98}\right) \\ &&-\frac12\left( \frac{1}{100} \right)\\\\ &=& \frac14-\frac12\left( \frac{1}{100} \right) \\\\ &=& \frac14- \frac{1}{200} \\\\ &\mathbf{=}& \mathbf{\frac{49}{200}} \\\\ &\mathbf{=}& \mathbf{0.245} \\ \hline \end{array}\)

 

laugh

 Dec 10, 2018
 #2
avatar+21855 
+11

Calculate:

\(\large{\frac{1}{11*15}+\frac{1}{19*15}+\frac{1}{19*23}+\frac{1}{23*27}+\cdots+\frac{1}{1103*1107}+\frac{1}{1111*1107}}\)

 

\(\begin{array}{|rcll|} \hline && \frac{1}{11*15}+\frac{1}{15*19}+\frac{1}{19*23}+\frac{1}{23*27}+\cdots+\frac{1}{1103*1107}+\frac{1}{1107*1111} \\\\ &=& \frac14\left( \frac{1}{11}-\frac{1}{15}\right) +\frac14\left( \frac{1}{15}-\frac{1}{19}\right) +\frac14\left( \frac{1}{19}-\frac{1}{23}\right) +\frac14\left( \frac{1}{23}-\frac{1}{27}\right) +\cdots\\ && +\frac14\left( \frac{1}{1103}-\frac{1}{1107}\right) +\frac14\left( \frac{1}{1107} -\frac{1}{1111} \right)\\\\ &=& \frac{1}{4*11} \\ &&-\frac14\left(\frac{1}{15}\right)+ \frac14\left(\frac{1}{15}\right)\\ &&-\frac14\left(\frac{1}{19}\right)+\frac14\left(\frac{1}{19}\right) \\ &&-\frac14\left(\frac{1}{23}\right)+\frac14\left(\frac{1}{23}\right) \\ &&-\frac14\left( \frac{1}{27}\right)+\frac14\left( \frac{1}{27}\right)+\cdots \\ &&-\frac14\left( \frac{1}{1103}\right)+\frac14\left( \frac{1}{1103}\right)\\ &&-\frac14\left( \frac{1}{1107}\right)+\frac14\left( \frac{1}{1107}\right) \\ &&-\frac14\left( \frac{1}{1111} \right)\\\\ &=& \frac{1}{44}-\frac14\left( \frac{1}{1111} \right) \\\\ &=& \frac{1}{44}- \frac{1}{4444} \\\\ &\mathbf{=}& \mathbf{\frac{275}{12221}} \\\\ &\mathbf{=}& \mathbf{\frac{25}{1111}} \\\\ &\mathbf{=}& \mathbf{0.\overline{0225}} \\ \hline \end{array}\)

 

laugh

 Dec 10, 2018
edited by heureka  Dec 11, 2018
 #3
avatar+60 
+1

Thanks, Heureka!

 

Note: 275/12221 can be simplified to 25/1111.

 Dec 10, 2018

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