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

   \(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2015^2}{2014.2016}\)

   \(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)

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

   \(=\frac{2015.2}{2016}\)

    \(=...\)(tự tinhs)

giúp mới mình sẽ tích

bn viết rõ đề đi

ai nhanh nhất mình sẽ cho 3 k

5435

tích mình nha

luu y : dau /la phan cach giua mau so va tu so

5435 tich nha

ket qua la 5435 ban nha

\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}\)

=\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)

=\(\frac{1}{2}\left(1-\frac{1}{2015}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2016}\right)\)

=\(\frac{3}{4}-\left(\frac{1}{4030}+\frac{1}{4032}\right)\) < \(\frac{3}{4}\)

=> đpcm

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

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

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

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

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

Vậy \(C=\frac{2015}{1008}\)

2015/2016

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