+0  
 
0
74
3
avatar+47 

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

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

 

 

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

 Dec 10, 2018
 #1
avatar+20831 
+6

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+20831 
+6

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+47 
+1

Thanks, Heureka!

 

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

 Dec 10, 2018

23 Online Users

avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.