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Compute \(\frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \frac{1}{4^3 - 4} + \dots + \frac{1}{100^3 - 100}\)

 Feb 28, 2019
 #1
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sumfor(n, 1, 100, (1/((n+1)^3 - (n+1)) = 0.2499514657

 Feb 28, 2019
 #2
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A slight adjustment in the answer above:

sumfor(n, 1, 99, (1/((n+1)^3 - (n+1)) = 0.249950495 

Guest Feb 28, 2019
 #3
avatar+23071 
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Compute

\(\mathbf{\large{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}}}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n^3-n} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n(n^2-1)} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{n(n-1)(n+1)} \\\\ &=& \sum \limits_{n = 2}^{100} \dfrac{1}{(n-1)n(n+1)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+1)(n+2)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+1)}\cdot \dfrac{1}{(n+2)} \quad | \quad \boxed{\dfrac{1}{n(n+1)} = \dfrac{1}{n} - \dfrac{1}{n+1}} \\\\ &=& \sum \limits_{n = 1}^{99} \left(\dfrac{1}{n} - \dfrac{1}{n+1} \right) \cdot \dfrac{1}{(n+2)} \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{n(n+2)} - \dfrac{1}{(n+1)(n+2)} \quad | \quad \boxed{\dfrac{1}{n(n+2)} = \dfrac{1}{2}\left( \dfrac{1}{n} - \dfrac{1}{n+2} \right) } \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n+2} \right) -\dfrac{1}{(n+1)(n+2)} \quad | \quad \boxed{\dfrac{1}{(n+1)(n+2)}= \dfrac{1}{n+1} - \dfrac{1}{n+2} } \\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2}\left(\dfrac{1}{n} - \dfrac{1}{n+2} \right) -\left( \dfrac{1}{n+1} - \dfrac{1}{n+2} \right)\\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{2(n+2)} - \dfrac{1}{n+1} + \dfrac{1}{n+2}\\\\ &=& \sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{n+2} \left(1 - \dfrac{1}{2} \right) \\\\ &\mathbf{=}& \mathbf{\sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &\mathbf{=}& \mathbf{\sum \limits_{n = 1}^{99} \dfrac{1}{2n} - \dfrac{1}{n+1} + \dfrac{1}{2(n+2)}} \\\\ &=& \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} \\\\ && \boxed{\dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n} \\ = \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 3}^{99} \dfrac{1}{n} \\ = \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} } \\ && \boxed{\sum \limits_{n = 1}^{99} \dfrac{1}{n+1} \\ = \dfrac12 + \sum \limits_{n = 2}^{99} \dfrac{1}{n+1} \\ = \dfrac12 + \sum \limits_{n = 1}^{98} \dfrac{1}{n+2} \\ = \dfrac12 + \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{100} } \\ && \boxed{ \dfrac{1}{2}\sum \limits_{n = 1}^{99} \dfrac{1}{n+2} \\ = \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} } \\\\ &=& \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\ && - \left( \dfrac12 + \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} + \dfrac{1}{100} \right) \\ && + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &=& \dfrac12+\dfrac14+\dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\ && - \dfrac12 - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \dfrac{1}{100} \\ && + \dfrac{1}{2}\sum \limits_{n = 1}^{97} \dfrac{1}{n+2} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101}\\ && +\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} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101}+ \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} - \sum \limits_{n = 1}^{97} \dfrac{1}{n+2} \\\\ &=& \dfrac14- \dfrac{1}{100} +\dfrac{1}{2\cdot 100}+\dfrac{1}{2\cdot 101} \\\\ &\mathbf{=}& \mathbf{\dfrac12 \left( \dfrac12- \dfrac{1}{50} +\dfrac{1}{100}+\dfrac{1}{101} \right)} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{100^3 - 100}} \\\\ &\mathbf{=}& \mathbf{\dfrac12 \left( \dfrac12- \dfrac{1}{50} +\dfrac{1}{100}+\dfrac{1}{101} \right)} \\\\ &\mathbf{=}& \dfrac{5049}{20200} \\\\ &\mathbf{=}& \mathbf{0.24995049505} \\ \hline \end{array}\)

 

laugh

 Mar 1, 2019
edited by heureka  Mar 1, 2019

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