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 #1
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Jan 19, 2018
 #6
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0

Compute the sum    2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5)+...

 

see link: http://web2.0calc.com/questions/algebra_47009

 

In General:

\(\begin{array}{lcll} s = \dfrac{1}{1 \cdot 2 } + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4 } + \dfrac{1}{4 \cdot 5 } + \cdots \ + \dfrac{1}{n \cdot (n+1)} + \cdots =\ \frac{1}{1!0!}\cdot \frac{1}{1} = 1 \\\\ s =\mathbf{ \dfrac{1}{1 \cdot 2 \cdot 3} + \dfrac{1}{2 \cdot 3 \cdot 4} + \dfrac{1}{3 \cdot 4 \cdot 5} + \dfrac{1}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \frac{1}{2!0!}\cdot \frac{1}{2} } = \frac{1}{4} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3)} + \cdots =\ \frac{1}{3!0!}\cdot \frac{1}{3} = \frac{1}{18} \\\\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7} + \dfrac{1}{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2) \cdot (n+3 \cdot (n+4) } + \cdots =\ \frac{1}{4!0!}\cdot \frac{1}{4} = \frac{1}{96} \\\\ \ldots \\ s = \dfrac{1}{1 \cdot 2 \cdot 3 \cdots } + \dfrac{1}{2 \cdot 3 \cdot 4 \cdots } + \dfrac{1}{3 \cdot 4 \cdot 5 \cdots} + \cdots \ + \dfrac{1}{n \cdot (n+1) \cdot (n+2)\cdot~\cdots ~\cdot (n+m) } + \cdots =\ \frac{1}{m!0!}\cdot \frac{1}{m} = \frac{1}{m\cdot m!} = \frac{1}{(m+1)!-m!} \\\\ \end{array} \\ \)

 

 

 

 

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Jan 19, 2018
 #1
avatar+28159 
+3
Jan 19, 2018

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