Ta có:\(P=\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)...\left(1-\dfrac{1}{1+2+...+2014}\right)\)
\(P=\dfrac{2}{1+2}\cdot\dfrac{2+3}{1+2+3}\cdot...\cdot\dfrac{2+3+...+2014}{1+2+3+...+2014}\)
\(P=\dfrac{\dfrac{1\cdot4}{2}}{\dfrac{2\left(2+1\right)}{2}}\cdot\dfrac{\dfrac{2\left(3+2\right)}{2}}{\dfrac{3\left(3+1\right)}{2}}\cdot...\cdot\dfrac{\dfrac{2013\left(2014+2\right)}{2}}{\dfrac{2014\left(2014+1\right)}{2}}\)
\(P=\dfrac{1\cdot4}{2\cdot3}\cdot\dfrac{2\cdot5}{3\cdot4}\cdot...\cdot\dfrac{2013\cdot2016}{2014\cdot2015}\)
\(P=\dfrac{1\cdot4\cdot2\cdot5\cdot...\cdot2013\cdot2016}{2\cdot3\cdot3\cdot4\cdot...\cdot2014\cdot2015}\)
\(P=\dfrac{\left(1\cdot2\cdot...\cdot2013\right)\left(4\cdot5\cdot...\cdot2016\right)}{\left(2\cdot3\cdot\cdot...\cdot2014\right)\left(3\cdot4\cdot...\cdot2015\right)}\)
\(P=\dfrac{2016}{2014\cdot3}\)
\(P=\dfrac{336}{1007}\)
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Câu hỏi của Phan Nguyễn Hà Linh - Toán lớp 6 - Học toán với OnlineMath
