\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+.....+\frac{1}{99\cdot100}\)
K
Khách
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QD
19 tháng 2 2017
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{1}{k}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)=\frac{1}{k}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Leftrightarrow\frac{1}{2}=\frac{1}{k}\Rightarrow k=2\)
ML
27 tháng 6 2015
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{n\left(n+2\right)}\)
\(\Rightarrow\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{1.2.3}+...+\frac{1}{98.99.100}=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(\Rightarrow k=2\)
26 tháng 5 2017
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
26 tháng 5 2017
Giải
A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/98-1/99+1/99-1/100
=1-1/100=99/100
Chú thích:1/2 là 1 phần 2
26 tháng 2 2018
Ta có 1/1.2-1/2.3=2/1.2.3;1/2.3-1/3.4=2/2.3.4 .....1/98.99-1/99.100=2/98.99.100 2A=2/1.2.3+2/2.3.4+....+2/98.99.100 = 1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100 = 1/2-1/99.100 = 4949/9900 A =4949/19800
MA
9
VA
28 tháng 7 2016
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
Ủng hộ mk nha !!! ^_^
28 tháng 7 2016
Ta có: \(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\\ =\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+.........+\left(\frac{1}{99}-\frac{1}{100}\right)\\ =\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+......+\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\\ =\frac{49}{100}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{99}.\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)