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Co quy luat nay ne em: 1+2=3=2.3:2; 1+2+3=6=3.4:2;...;1+2+3+...+2012=2012.2013:2
Suy ra ta co:
Mau so cua D=1 + 1/(2.3:2) + 1/(3.4:2) + 1/(4.5:2) + .... + 1/(2012.2013:2)
=1 + 2/2.3 + 2/3.4 + 2/4.5 + .... + 2/2012.2013
= 2.[1/2 + 1/2.3 + 1/3.4 + 1/4.5 + .... + 1/2012.2013]
=2.[1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ..... + 1/2012.2013]
=2.[1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +....+1/2012 - 1/2013
=2[1 - 1/2013]
=2.2012/2013
Vay D= 2.2012 / (2.2012:2013)=2013
\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).....\left(1-\frac{1}{1+2+3+.....+2018}\right)\)
\(=\left(1-\frac{1}{\frac{2.3}{2}}\right).\left(1-\frac{1}{\frac{3.4}{2}}\right).......\left(1-\frac{1}{\frac{2018.2019}{2}}\right)\)
\(=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).......\left(1-\frac{2}{2018.2019}\right)\)
\(=\left(1-\frac{1}{3}\right).\left(1-\frac{5}{6}\right).......\left(1-\frac{1}{2037171}\right)\)
\(=\frac{2}{3}.\frac{5}{6}......\frac{2037170}{2037171}\)
\(=\frac{4}{6}.\frac{10}{12}.......\frac{4074340}{4074342}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}......\frac{2017.2020}{2018.2019}\)
\(=\frac{1.2......2017}{2.3.....2018}.\frac{4.5......2020}{3.4......2019}=\frac{1}{2018}.\frac{2020}{3}=\frac{1010}{3027}\)
\(D=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}\right):\left(\frac{2011}{1}+\frac{2010}{2}+...+\frac{1}{2011}\right)\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)+1}\)
\(\Rightarrow D=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}}\)
\(\Rightarrow D\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{2012\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)}\)
\(\Rightarrow D=\frac{1}{2012}\)
Vũ Minh Tuấn giúp mình vs