tính m=(1+1/1.3)+(1+1/2.4)+(1+3.5).....(1+1/2014.2016)
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\(C=\left(1+\frac{1}{1.3}\right)\)\(.\left(1+\frac{1}{2.4}\right)\)\(.\left(1+\frac{1}{3.5}\right)\)\(.\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2015^2}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)
\(=\frac{\left(2.3.4...2015\right).\left(2.3.4...2015\right)}{\left(1.2.3...2014\right).\left(3.4.5...2016\right)}\)
\(=\frac{2015.2}{2016}\)
\(=...\)(tự tinhs)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(=\frac{4}{3}.\frac{9}{8}...\frac{4060225}{4060224}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3...2015.2015}{1.3.2.4...2014.2016}\)
\(=\frac{2.3...2015}{1.2...2014}.\frac{2.3...2015}{3.4...2016}\)
\(=2015.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}\)
\(=\frac{2015}{1008}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4.3.5...2014.2016}\)
\(=\frac{\left(2.3.4..2015\right)\left(2.3.4..2015\right)}{\left(1.2.3..2014\right)\left(3.4.5..2016\right)}\)
\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)
Vậy \(C=\frac{2015}{1008}\)
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