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\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\)
\(\Rightarrow A=\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+...+\frac{1}{19}.\frac{1}{20}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2015}\)
\(=\frac{1}{\left(1+0\right).2:2}+\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+...+\frac{1}{\left(1+2015\right).2015:2}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2015.2016}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}\right)\)
\(=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=2.\left(1-\frac{1}{2016}\right)\)
\(=2.\frac{2015}{2016}=\frac{2015}{1008}\)
1-\(\frac{1}{2}\)+ \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{2013}\)- \(\frac{1}{2014}\)
=(1+\(\frac{1}{3}\)+...+\(\frac{1}{2013}\)) - (\(\frac{1}{2}\)+ \(\frac{1}{4}\) + ...+ \(\frac{1}{2014}\))
=(1+\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+ \(\frac{1}{2013}\)+ \(\frac{1}{2014}\))-2.(\(\frac{1}{2}\)+ \(\frac{1}{4}\)+...+\(\frac{1}{2014}\))
=1+\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+ \(\frac{1}{2013}\)+ \(\frac{1}{2014}\)- 1-\(\frac{1}{2}\)-...-\(\frac{1}{1007}\)
=(1+\(\frac{1}{2}\)+ \(\frac{1}{3}\)+ \(\frac{1}{4}\)+...+\(\frac{1}{1007}\))+\(\frac{1}{1008}\)+ \(\frac{1}{1009}\)+...+\(\frac{1}{2013}\)+ \(\frac{1}{2014}\)-(1+\(\frac{1}{2}\)+...+\(\frac{1}{1007}\))
=\(\frac{1}{1008}\)+ \(\frac{1}{1009}\)+...+\(\frac{1}{2013}\)+ \(\frac{1}{2014}\).
mình chưa hiểu lắm
tại sao nhân 2 lên và còn 1 - \(\frac{1}{2}\)- ... - \(\frac{1}{1007}\)
1007 ở đâu?????
Ta có: \(\frac{\frac{2014}{1}+\frac{2013}{2}+\frac{2012}{3}+...\frac{1}{2014}+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\left(\frac{2013}{2}+1\right)+\left(\frac{2012}{3}+1\right)+...+\left(\frac{1}{2014}+1\right)+1+2014}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=
= \(\frac{\frac{2015}{2}+\frac{2015}{3}+...+\frac{2015}{2014}+2015}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=\(\frac{2015.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}+1\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)=2015
\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2015}\right)\)
\(=\left(\frac{2-1}{2}\right)\left(\frac{3-1}{3}\right)\left(\frac{4-1}{4}\right)....\left(\frac{2015-1}{2015}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2013}{2014}.\frac{2014}{2015}\)
\(=\frac{1}{2015}\)
2015/2 đó bạn, tick cho mình nhé