+0  
 
-1
101
2
avatar+274 

Calculate

\(\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{99\cdot100\cdot101}\)

 

\((A) \frac {5029}{20200}\) \((B) \frac{5039}{20200}\) \((C) \frac {3049}{20200}\) \((D) \frac {509}{20200}\) \((E) \frac {5049}{20200}\)

Thanks! 

 Mar 9, 2020
 #1
avatar+25235 
+1

Calculate

\(\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{2\cdot3\cdot4}+\ldots+\dfrac{1}{99\cdot100\cdot101}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{2\cdot3\cdot4}+\ldots+\dfrac{1}{99\cdot100\cdot101} } \\ &=& \sum \limits_{n=1}^{99} \dfrac{1}{n(n+1)(n+2)} \\ &=& \sum \limits_{n=1}^{99} \dfrac{1}{n}\left( \dfrac{1}{(n+1)(n+2)} \right) \\ &=& \sum \limits_{n=1}^{99} \dfrac{1}{n}\left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\ &=& \sum \limits_{n=1}^{99} \left(\dfrac{1}{n(n+1)}-\dfrac{1}{n(n+2)} \right) \\ &=& \sum \limits_{n=1}^{99} \left(\dfrac{1}{n}-\dfrac{1}{n+1} - \dfrac{1}{2}\left(\dfrac{1}{n}-\dfrac{1}{n+2} \right) \right) \\ &=& \sum \limits_{n=1}^{99} \left(\dfrac{1}{2n}-\dfrac{1}{n+1} + \dfrac{1}{2(n+2)} \right) \\ &=& \mathbf{\sum \limits_{n=1}^{99} \dfrac{1}{2n} -\sum \limits_{n=1}^{99} \dfrac{1}{n+1} +\sum \limits_{n=1}^{99} \dfrac{1}{2(n+2)} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \sum \limits_{n=1}^{99} \dfrac{1}{2n} -\sum \limits_{n=1}^{99} \dfrac{1}{n+1} +\sum \limits_{n=1}^{99} \dfrac{1}{2(n+2)} \\\\ &=& \dfrac{1}{2} \sum \limits_{n=1}^{99} \dfrac{1}{n} -\sum \limits_{n=1}^{99} \dfrac{1}{n+1} +\dfrac{1}{2}\sum \limits_{n=1}^{99} \dfrac{1}{n+2} \\\\ &=& \dfrac{1}{2} \sum \limits_{n=-1}^{97} \dfrac{1}{n+2} -\sum \limits_{n=0}^{98} \dfrac{1}{n+2} +\left( \dfrac{1}{2}* \dfrac{1}{100} + \dfrac{1}{2}* \dfrac{1}{101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \left( \dfrac{1}{2}* \dfrac{1}{1} +\dfrac{1}{2}*\dfrac{1}{2} + \dfrac{1}{2} \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\ && -\sum \limits_{n=0}^{98} \dfrac{1}{n+2} \\ && +\left( \dfrac{1}{2*100} + \dfrac{1}{2*101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \left( \dfrac{1}{2}+ \dfrac{1}{4} + \dfrac{1}{2} \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\ && - \dfrac{1}{2} - \sum \limits_{n=1}^{98} \dfrac{1}{n+2} \\ && +\left( \dfrac{1}{2*100} + \dfrac{1}{2*101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \left( \dfrac{1}{2}+ \dfrac{1}{4} + \dfrac{1}{2} \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\ && - \dfrac{1}{2} - \dfrac{1}{100} - \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \\ && +\left( \dfrac{1}{2*100} + \dfrac{1}{2*101}+ \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \right) \\\\ &=& \dfrac{1}{2}+ \dfrac{1}{4} - \dfrac{1}{2} - \dfrac{1}{100}+\dfrac{1}{200} + \dfrac{1}{202} \\ && + \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} - \sum \limits_{n=1}^{97} \dfrac{1}{n+2} + \dfrac{1}{2}\sum \limits_{n=1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{100}+ \dfrac{1}{200} + \dfrac{1}{202} + \sum \limits_{n=1}^{97} \dfrac{1}{n+2} - \sum \limits_{n=1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{100}+ \dfrac{1}{200} + \dfrac{1}{202} \\\\ &=& \dfrac{1}{4} - \dfrac{2}{200}+ \dfrac{1}{200} + \dfrac{1}{202} \\\\ &=& \dfrac{1}{4} - \dfrac{1}{200} + \dfrac{1}{202} \\\\ &=& \dfrac{1}{2} \left( \dfrac{1}{2} - \dfrac{1}{100} + \dfrac{1}{101} \right) \\\\ &=& \dfrac{1}{2} \left( \dfrac{98}{200} + \dfrac{1}{101} \right) \\\\ &=& \dfrac{1}{2} \left( \dfrac{98*101+200}{200*101} \right) \\\\ &=& \dfrac{49*101+100}{200*101} \\\\ &=& \mathbf{ \dfrac{5049}{20200} } \\ \hline \end{array}\)

 

laugh

 Mar 12, 2020
 #2
avatar
0

∑[ 1 / ( n^3 + 3 n^2 + 2 n), n, 1, 99] =5049 / 20200

 Mar 12, 2020

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