Summer Clouds moderators
04/07/2017 at 09:17-
Consider the following expression :
\(\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}\times\dfrac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\dfrac{1}{2}.\dfrac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}=\dfrac{1}{2}.\left(\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
Denote \(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{997.998.999}+\dfrac{1}{998.999.1000}\)
\(\Rightarrow A=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)+\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)+...+\dfrac{1}{2}\left(\dfrac{1}{997.998}-\dfrac{1}{998.999}\right)+\dfrac{1}{2}\left(\dfrac{1}{998.999}-\dfrac{1}{999.1000}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{997.998}-\dfrac{1}{998.999}+\dfrac{1}{998.999}-\dfrac{1}{999.1000}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{999000}\right)=\dfrac{1}{4}-\dfrac{1}{1998000}=\dfrac{499499}{1998000}\)
Selected by MathYouLike -
My Name Does Not Have To Ask 05/07/2017 at 14:25\(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{998\cdot999\cdot1000}\)
Apply formula : \(\dfrac{2n}{a\left(a+n\right)\left(a+2n\right)}=\dfrac{1}{a\left(a+n\right)}-\dfrac{1}{\left(a+n\right)\left(a+2n\right)}\)
\(2.B=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{998\cdot999\cdot1000}\)
\(2.B=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{998\cdot999}-\dfrac{1}{999\cdot1000}\)
\(2.B=\dfrac{1}{1\cdot2}-\dfrac{1}{999\cdot1000}\)
\(2B=\dfrac{499499}{999000}\)
\(\Rightarrow B=\dfrac{499499}{1998000}\)
