A=2016.(1+1/1.3).(1+1/2.4).(1+1/3.5)...(1+1/2014.2016)
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Tính
B=2016.\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)............\left(1+\frac{1}{2014.2016}\right)\)
\(B=2016.\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
= \(2016.\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}....\frac{2015^2}{2014.2016}\)
= \(2016.\frac{2.3.4....2015}{1.2.3.4.5...2014.2015.2016}.\frac{2.3.4....2015}{3.4.5...2014}\)
= \(2016.\frac{1}{2016}.2.2015=2.2015=4030\)
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C=(1+1/1.3)(1+1/2.4)(1+1/3.5)......(1+1/2014.2016)
\(C=\left(1+\frac{1}{1.3}\right)\)\(.\left(1+\frac{1}{2.4}\right)\)\(.\left(1+\frac{1}{3.5}\right)\)\(.\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}...\frac{2015^2}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)
\(=\frac{\left(2.3.4...2015\right).\left(2.3.4...2015\right)}{\left(1.2.3...2014\right).\left(3.4.5...2016\right)}\)
\(=\frac{2015.2}{2016}\)
\(=...\)(tự tinhs)
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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\(\frac{\left(1.3+1\right)\left(2.4+1\right)\left(3.5+1\right).......\left(2014.2016+1\right)}{1.3.2.4.3.5...............2014.2016}\)
A\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)........\left(1+\frac{1}{2014.2016}\right)\)
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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\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
Có \(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)..........\)\(\left(1+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)....\left(\frac{2014.2016}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\left(\frac{2^2-1}{1.3}+\frac{1}{2.4}\right)\left(\frac{3^2-1}{2.4}+\frac{1}{2.4}\right)......\left(\frac{2015^2-1}{2014.2016}+\frac{1}{2014.2016}\right)\)
=\(\frac{2.2}{1.3}.\frac{3.3}{2.4}......\frac{2015.2015}{2014.2016}\)
=\(\frac{2.2.3.3.....2015.2015}{1.3.2.4....2014.2015}\)
=\(\frac{\left(2.3...2015\right).\left(2.3.....2015\right)}{\left(1.2....2014\right).\left(3.4.....2016\right)}=\frac{2015.2}{2016}=\frac{4030}{2016}\)