Violympic toán 6

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Cho A=\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.....+\dfrac{1}{x.\left(x+1\right)}\)\(=\dfrac{2014}{2016}\)

Lightning Farron
14 tháng 3 2017 lúc 21:35

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2014}{2016}\)

\(A=\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2014}{2016}\)

\(A=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2014}{2016}\)

\(A=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{1007}{2016}\)

\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)

\(A=\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1007}{2016}\)

\(A=\dfrac{1}{x+1}=\dfrac{1}{2016}\)\(\Leftrightarrow x+1=2016\Leftrightarrow x=2015\)