C=(1+1/1.3)(1+1/2.4)(1+1/3.5)......(1+1/2014.2016)
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c= (1+1/1.3).(1+1/2.4).(1+1/3.5)*...*(1+1/2014.2016)
tính m=(1+1/1.3)+(1+1/2.4)+(1+3.5).....(1+1/2014.2016)
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tính :(1+1/1.3).(1+1/2.4)+(1+1/3.5)+...+(1+1/2014.2016)
luu y : dau /la phan cach giua mau so va tu so
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A=2016.(1+1/1.3).(1+1/2.4).(1+1/3.5)...(1+1/2014.2016)
\(\frac{\left(1.3+1\right)\left(2.4+1\right)\left(3.5+1\right).......\left(2014.2016+1\right)}{1.3.2.4.3.5...............2014.2016}\)
Tính C:
C = ( 1+1/1.3)(1+1/2.4)(1+1/3.5).....(1+1/2014.2016)
(Mình đang rất vội nên mong các bạn trả lời sớm. Cảm ơn nhiều!)
Chứng minh rằng: 1/1.3 + 1/2.4 + 1/3.5 + 1/4.6 + ...+ 1/2013.2015 + 1/2014.2016 < 3/4
Giúp!!!!!!!!
\(\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+...+\frac{1}{2013.2015}+\frac{1}{2014.2016}\)
=\(\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2013.2015}\right)+\left(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2014.2016}\right)\)
=\(\frac{1}{2}\left(1-\frac{1}{2015}\right)+\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2016}\right)\)
=\(\frac{3}{4}-\left(\frac{1}{4030}+\frac{1}{4032}\right)\) < \(\frac{3}{4}\)
=> đpcm
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Tính:
C=\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
\(C=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{2015.2015}{2014.2016}\)
\(=\frac{2.2.3.3.4.4...2015.2015}{1.3.2.4.3.5...2014.2016}\)
\(=\frac{\left(2.3.4..2015\right)\left(2.3.4..2015\right)}{\left(1.2.3..2014\right)\left(3.4.5..2016\right)}\)
\(=\frac{2015.2}{2016}=\frac{2015}{1008}\)
Vậy \(C=\frac{2015}{1008}\)
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C=\(\frac{2015}{1008}\)
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tinh
C=\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)+....+\left(1+\frac{1}{2014.2016}\right)\)