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Compute \(\dfrac 34+\dfrac3{28}+\dfrac3{70}+\dfrac3{130}+\cdots +\dfrac3{9700}\)

 

Note: The denominators follow a quadratic function.

 Jun 19, 2020
 #1
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sum_(n=1)^33 3/((3 n - 2) (3 n + 1)) = 99/100

 Jun 19, 2020
 #2
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Compute


\(\dfrac 34+\dfrac3{28}+\dfrac3{70}+\dfrac3{130}+\cdots +\dfrac3{9700}\)

 

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac 34+\dfrac3{28}+\dfrac3{70}+\dfrac3{130}+\cdots +\dfrac3{8554}+\dfrac3{9118}+\dfrac3{9700}} \\\\ &=& \sum \limits_{k=0}^{32} \dfrac{3}{(3k+1)(3k+4)} \\\\ &=& \sum \limits_{k=0}^{32}\left( \dfrac{1}{3k+1} -\dfrac{1}{3k+4} \right) \\\\ &=& \sum \limits_{k=0}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0}^{32}\left(\dfrac{1}{3k+4}\right) \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0}^{31}\left(\dfrac{1}{3k+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0+1}^{31+1}\left(\dfrac{1}{3(k-1)+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=1}^{32}\left(\dfrac{1}{3k-3+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \dfrac{1}{100} \\\\ &=& 1 - \dfrac{1}{100} \\\\ &=& \dfrac{100-1}{100} \\\\ &=& \mathbf{\dfrac{99}{100}} \\ \hline \end{array}\)

 

laugh

 Jun 19, 2020
edited by heureka  Jun 19, 2020

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