Question

Tính:

C=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)

Answer 1

\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..\left(1+\frac{1}{2014.2016}\right)\)

\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2015.2015}{2014.2016}\)

\(=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4.3.5...2014.2016}\)

\(=\frac{\left(2.3.4..2015\right)\left(2.3.4..2015\right)}{\left(1.2.3..2014\right)\left(3.4.5..2016\right)}\)

\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)

Vậy \(C=\frac{2015}{1008}\)

Answer 2

2015/2016

Answer 3

C=\(\frac{2015}{1008}\)

Answer 4

trả lời

\(C=\frac{2005}{1008}\)

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