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\(A=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}\times...\times\frac{2015}{2016}\times\frac{2016}{2017}=\frac{1}{2017}\)
=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.......+\(\frac{1}{2016}\)-\(\frac{1}{2017}\)+1
=\(\frac{1}{1}\)-\(\frac{1}{2017}\)+1
=\(\frac{2016}{2017}\)+1
=\(\frac{1}{2017}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}+1\)
\(=1-\frac{1}{2017}+1\)
\(=\frac{2016}{2017}+1\)
\(=\frac{4033}{2017}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2016}-\frac{1}{2017}.\)
=> \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2016.2017}=\frac{1}{2}-\frac{1}{2017}< \frac{1}{2}\)
\(B=\)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)..\left(1-\frac{1}{2016}\right)\left(1-\frac{1}{2017}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2015}{2016}.\frac{2016}{2017}\)
\(\Rightarrow B=\frac{1}{2017}\)
Ta có:\(B=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times............\times\left(1-\frac{1}{2017}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times............\times\frac{2016}{2017}\)
\(=\frac{1\times2\times..........\times2016}{2\times3\times...........\times2017}=\frac{1}{2017}\)
1+2+.....+100 = 5050
2+4+....+2016 = 1017072
1+3+5+....+2017=1018081
A=(1-1/2)x(1-1/3)x.................x(1-1/2017)
A=\(\frac{1}{2}\times\frac{2}{3}\)x....x\(\frac{2016}{2017}\)
A=\(\frac{1}{2017}\)
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{2}{3}\right)...\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}.\frac{2}{3}...\frac{2015}{2016}.\frac{2016}{2017}\)
\(A=\frac{1.2.3...2016}{2.3.4...2017}=\frac{1}{2017}\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2016}\right)\cdot\left(1-\frac{1}{2017}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2015}{2016}\cdot\frac{2016}{2017}\)
\(A=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2015\cdot2016}{2\cdot3\cdot4\cdot5\cdot....\cdot2016\cdot2017}\)
\(A=\frac{1}{2017}\)