c= (1+1/1.3).(1+1/2.4).(1+1/3.5)*...*(1+1/2014.2016)
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TN
1
6 tháng 4 2022
\(C=\left(1+\frac{1}{1.3}\right)\)\(.\left(1+\frac{1}{2.4}\right)\)\(.\left(1+\frac{1}{3.5}\right)\)\(.\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2015^2}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)
\(=\frac{\left(2.3.4...2015\right).\left(2.3.4...2015\right)}{\left(1.2.3...2014\right).\left(3.4.5...2016\right)}\)
\(=\frac{2015.2}{2016}\)
\(=...\)(tự tinhs)
CC
10
V
9
ND
0
NM
1
KB
21 tháng 5 2017
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}\)
=\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{2015}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2016}\right)\)
=\(\frac{3}{4}-\left(\frac{1}{4030}+\frac{1}{4032}\right)\) < \(\frac{3}{4}\)
=> đpcm
HB
20 tháng 4 2018
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4.3.5...2014.2016}\)
\(=\frac{\left(2.3.4..2015\right)\left(2.3.4..2015\right)}{\left(1.2.3..2014\right)\left(3.4.5..2016\right)}\)
\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)
Vậy \(C=\frac{2015}{1008}\)