A= (1-1/2)x(1-1/3)x(1-1/4)x.......x (1-1/2013)x(1-1/2014)=?
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\(A=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x......x\left(1-\frac{1}{2013}\right)x\left(1-\frac{1}{2014}\right)\)
\(A=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...............x\frac{2012}{2013}x\frac{2013}{2014}\)
\(A=\frac{1}{2014}\)
\(\left[1-\frac{1}{2}\right]\left[1-\frac{1}{3}\right]...\left[1-\frac{1}{2014}\right]\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}...\cdot\frac{2013}{2014}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2013}{2\cdot3\cdot4\cdot5\cdot...\cdot2014}=\frac{1}{2014}\)
A=1/2*2/3*3/4*...*2012/2013*2013/2014
Ta gạch bỏ các chữ số giống nhau còn lại 1/2014
Vậy A= 1/2014
\(a)\) \(S=1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{2187}\)
\(S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\)
\(3S=3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\)
\(3S-S=\left(3+1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^6}\right)-\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^7}\right)\)
\(2S=3+\frac{1}{3^7}\)
\(2S=\frac{3^8+1}{3^7}\)
\(S=\frac{3^8+1}{3^7}.\frac{1}{2}\)
\(S=\frac{3^8+1}{2.3^7}\)
Vậy \(S=\frac{3^8+1}{2.3^7}\)
Chúc bạn học tốt ~
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot...\cdot\frac{4056195}{2014\cdot2014}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}\)
\(-A=\frac{1\cdot2015}{2014\cdot2}=\frac{2015}{4028}\)
\(A=\frac{-2015}{4028}\)