Đặt \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{2}-\frac{1}{x+1}\)
\(\Rightarrow A=\left(\frac{1}{2}-\frac{1}{x+1}\right):\frac{1}{2}\)
Theo bài ra ta có:
\(\left(\frac{1}{2}-\frac{1}{x+1}\right):\frac{1}{2}=\frac{2011}{2013}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2011}{2013}.\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{2013}{4026}-\frac{2011}{4026}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2013}\)
=> x + 1 = 2013
=> x = 2013 - 1
=> x = 2012 \(\in\) N
Vậy x = 2012
Đặt S=1/3+1/6+1/10+..........+2/x(x+1)
1/2S=1/2[1/3+1/6+1/10+...+2/x(x+1)]
1/2S=1/6+1/12+1/20+......1/x(x+1)
1/2S=1/2.3+1/3.4+1/4.5+.....+1x(x+1)
1/2S=1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/x+1
1/2S=1/2-1/x+1
Vì S=2011/2013
suy ra (1/2-1/x+1):1/2=2011/2013
(1/2-1/x+1).2=2011/2013
1/2-1/x+1=2011/2013:2
1/2-1/x+1=2011/4026
1/x+1=1/2-2011/4026
1/x+1=1/2013
suy ra x+1=2013
x=2013-1
x=2012
