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Ta có :
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2004.2005}\)
\(=\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2004}-\frac{1}{2005}\)
\(=\)\(\frac{1}{2}-\frac{1}{2005}\)
\(=\)\(\frac{2005}{4010}-\frac{2}{4010}\)
\(=\)\(\frac{2003}{4010}\)
Chúc bạn học tốt ~
Gọi \(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2014\cdot2015}\)
\(A=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{2014\cdot2015}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}\)
\(A=\frac{1}{2}-\frac{1}{2015}\)
\(A=\frac{2015}{4030}-\frac{2}{4030}\)
\(A=\frac{2013}{4030}\)
\(1.2.3....2015-1.2.3....2014-1.2.3....2013.2014^2\)
\(=1.2.3...\left(2014+1\right)-1.2.3...\left(2014+1\right)\)
\(=0\)
a/ -1-2-3-...-2011-2017
= - (1+2+3+...+2017)
\(=-\frac{2017.2018}{2}=-2035153\)
b/ 1+(-2)+3+(-4)+...+2015+(-2016)
= (1-2)+(3-4)+...+(2015-2016)
= -1-1-...-1=-1008
c/ 2-4-6+8+10-12-14+16+...+1994-1996-1998+2000
= (2-4-6+8)+(10-12-14+16)+...+(1994-1996-1998+2000)
= 0+0+...+0 = 0
Đặt A= \(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\)
=>\(\frac{1}{4}A=\frac{1}{4}\left(\frac{4}{1.2}+\frac{4}{2.3}+...+\frac{4}{2014.2015}\right)\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)
=\(1-\frac{1}{2015}\)
=\(\frac{2014}{2015}\)
vậy ....
dấu "." là dấu nhân
\(\frac{4}{1\cdot2}+\frac{4}{2\cdot3}+\frac{4}{3\cdot4}+...+\frac{4}{2014\cdot2015}\)
\(=4\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\right)\)
\(=4\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=4\left(1-\frac{1}{2015}\right)\)
\(=4\cdot\frac{2014}{2015}\)
\(=\frac{8056}{2015}\)