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\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{100.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)\(=\frac{101}{101}-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2018\cdot2019}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(A=1-\frac{1}{2019}=\frac{2018}{2019}\)
Mà \(\frac{2018}{2019}< \frac{2019}{2019}=1\)
\(\Rightarrow A< 1\)
=\(11\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99\cdot100}\right)\)=\(11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)=\(11\left(1-\frac{1}{100}\right)\)=11\(\frac{99}{100}\)=\(\frac{1089}{100}\)
Đặt \(A=\frac{11}{1.2}+\frac{11}{2.3}+...+\frac{11}{99.100}\)
\(\Rightarrow A=11\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Rightarrow A=11\left(1-\frac{1}{100}\right)\)
\(\Rightarrow A=11.\frac{99}{100}\)
\(\Rightarrow A=\frac{1089}{100}\)
A=1.2+2.3+3.4+........+98.99
3A=1.2.3+2.3.3+3.4.3+........+98.99.3
3A=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97)
3A=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99
3A=98.99.100
===>A=(98.99.100)/3
#Japhkiel#
A=1.2+2.3+3.4+........+98.99
3A=1.2.3+2.3.3+3.4.3+........+98.99.3
3A=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97)
3A=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99
3A=98.99.100
A=\(\frac{98.99.100}{3}=\frac{970200}{3}=323400\)
(1-1/1.2)+(1-1/2*3)+......+(1-1/2015*2016)
=(0/1*2)+(0+2*3)+..........+(0/2015*2016)
=0
tui nghĩ cái đề phải như thế này \(\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{2015.2016}\right)\)
\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2015}\)
\(=\frac{2}{1.2}+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+2+3\right).3}{2}}+.....+\frac{1}{\frac{\left(2015+1\right).2015}{2}}\)
\(=\frac{2}{1.2}+\frac{2}{2.3}+....+\frac{2}{2015.2016}\)
1/1.2+1/2.3+1/3.4+......+1/2003.2004=1/1-1/2+1/2-1/3+1/3-1/4+......+1/2003-1/2004
=1/1-1/2004
=2003/2004
1/1.2+1/2.3+1/3.4+.......1/2003.2004
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2003}-\frac{1}{2004}\)
\(=1-\frac{1}{2004}\)
\(=\frac{2003}{2004}\)
Đặt \(A=\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+...+\frac{7}{10\cdot11}\)
\(\Rightarrow\frac{1}{7}A=\frac{1}{7}\left(\frac{7}{1\cdot2}+\frac{7}{2\cdot3}+...+\frac{7}{10\cdot11}\right)\)
\(\Rightarrow\frac{1}{7}A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\)
\(\Rightarrow\frac{1}{7}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow\frac{1}{7}A=1-\frac{1}{11}\)
\(\Rightarrow\frac{1}{7}A=\frac{10}{11}\)
\(\Rightarrow A=\frac{10}{11}:\frac{1}{7}\)
\(\Rightarrow A=\frac{70}{11}\)