Evaluate the following sum
$$\frac {1}{1×4×7} +\frac {1}{4×7×10}+\frac {1}{7×10×13}+....\frac {1}{25×28×31}$$
How could i solve this problem?
Thanks very much for help
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Does it telescope? – Angina Seng Dec 03 '17 at 19:06
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What does "telescope" mean? – user373141 Dec 03 '17 at 19:08
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1A sum $\sum a_n$ telescopes if there is a convenient sequence $b_n$ with $a_n=b_n-b_{n+1}$ for all $n$. – Angina Seng Dec 03 '17 at 19:17
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Related since the techniques in the following link may also be applied here: Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$ – Simply Beautiful Art Dec 04 '17 at 00:50
4 Answers
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Hint. Try to decompose each fraction like this: $$\frac{1}{(n-3) \cdot n \cdot (n+3)} = \frac{1}{18} \left( \frac{1}{n-3} - \frac{2}{n} + \frac{1}{n+3} \right) $$
Anastasia K.
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Hint: $$\frac { 1 }{ a\left( a+3 \right) \left( a+6 \right) } =\frac { 1 }{ 6 } \left[ \frac { 1 }{ a\left( a+3 \right) } -\frac { 1 }{ \left( a+3 \right) \left( a+6 \right) } \right] =\frac { 1 }{ 6 } \left[ \frac { 1 }{ 3 } \left( \frac { 1 }{ a } -\frac { 1 }{ a+3 } \right) -\frac { 1 }{ 3 } \left( \frac { 1 }{ a+3 } -\frac { 1 }{ a+6 } \right) \right] =\\ =\frac { 1 }{ 18 } \left[ \frac { 1 }{ a } -\frac { 2 }{ a+3 } +\frac { 1 }{ a+6 } \right] $$
haqnatural
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Partial fractions gives \begin{eqnarray*} \frac{1}{n(n+3)(n+6) } =\frac{1}{18} \left( \frac{1}{n}-\frac{2}{n+3}+\frac{1}{n+6} \right) \end{eqnarray*} so the sum does telescope \begin{eqnarray*} \frac{1}{18} \left[ \left(\frac{1}{1}-\frac{2}{4}+\frac{1}{7} \right)+ \left(\frac{1}{4}-\frac{2}{7}+\frac{1}{10} \right)+\cdots+\left(\frac{1}{22}-\frac{2}{25}+\frac{1}{28} \right)+\left(\frac{1}{25}-\frac{2}{28}+\frac{1}{31} \right)\right] \\ \frac{1}{18} \left( \frac{1}{1}- \frac{1}{4}-\frac{1}{28}+\frac{1}{31}\right)=\color{red}{\frac{9}{217}}. \end{eqnarray*}
Donald Splutterwit
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One does the trick like the following:
\begin{align*} \sum_{n}\left(\dfrac{1}{n}-\dfrac{2}{n+3}+\dfrac{1}{n+6}\right)&=\sum_{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)+\sum_{n}\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)\\ &~~~~+\sum_{n}\left(\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)+\sum_{n}\left(-\dfrac{1}{n+3}+\dfrac{1}{n+4}\right)\\ &~~~~+\sum_{n}\left(\dfrac{1}{n+5}-\dfrac{1}{n+4}\right)+\sum_{n}\left(-\dfrac{1}{n+5}+\dfrac{1}{n+6}\right) \end{align*} and finally telescoping takes place.
user284331
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