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2c=2/1.3+2/2.4+2/3.5+...+2/2016.2018
2c=1-1/3+1/2-1/4+1/3-1/5+...+1/2016-1/2017
2c=1-1/2017
2c=2016/2017
c=4032/2017
Lê Minh Hiếu cách lm đúng mà sai kết quả
KQ : \(\frac{2016}{4034}\)
I: Để 3n+4/n+2 là số nguyên thì \(3n+4⋮n+2\)
\(\Leftrightarrow3n+6-2⋮n+2\)
\(\Leftrightarrow n+2\in\left\{1;-1;2;-2\right\}\)
hay \(n\in\left\{-1;-3;0;-4\right\}\)
II: \(D=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)\)
\(D=2\cdot\left(1-\dfrac{1}{2009}\right)=2\cdot\dfrac{2008}{2009}=\dfrac{4016}{2009}\)
=1/2.(1+1/1.3).(1+1/2.4).(1+1/3.5)...(1+1/2014.2016)
=1/2.(1+1/1-1/3).(1+1/3-1/5)...(1+1/2014-1/2016)
=1/2.1+(1/1-1/2016)
=1/2.2015/2016
=2015/4032
\(A=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{2014.2016}\right)\)
\(A=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{2015.2015}{2014.2016}\)
\(A=\frac{2.3.4...2015}{1.2.3...2014}.\frac{2.3.4...2015}{3.4.5...2016}\)
\(A=2015.\frac{1}{1008}\)
\(A=\frac{2015}{1008}\)
Ta có :
\(A=\frac{2^2}{1.3}.\frac{3^2}{2.4}............\frac{2015^2}{2014.2016}\)= \(\frac{2.2}{1.3}.\frac{3.3}{2.4}...........\frac{2015.2015}{2014.2016}=\frac{2.2015}{2016}=\frac{2015}{1008}\)
k cho mình nha