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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
Ta có: \(A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+\dfrac{1}{3\cdot4\cdot5}+...+\dfrac{1}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+\dfrac{2}{3\cdot4\cdot5}+...+\dfrac{2}{98\cdot99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}-\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}-\dfrac{1}{4\cdot5}+...-\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=-\dfrac{1}{2}+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow2A=\dfrac{-1}{2}+\dfrac{1}{9900}\)
\(\Leftrightarrow2A=\dfrac{-4950}{9900}+\dfrac{1}{9900}=\dfrac{-4949}{9900}\)
hay \(A=\dfrac{-4949}{19800}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=1/2-1/3-1/4+1/3-1/4-1/5+1/5-1/6-1/7+...+1/35-1/36-1/37
giao hoán, kết hợp là ra nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giúp mình với. Cảm ơn nhiều!
Tính:
1/(1x2x3) + 1/(2x3x4) + 1/(3x4x5) + ... + 1/(100x101x102) = ?
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{100.101.102}\)
\(\Rightarrow\frac{1}{2}A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{100.101.102}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\)
\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{101.102}=\frac{2575}{5151}\Leftrightarrow A=\frac{2575}{10302}\)
ta co:1/1*2*3=(1/1*2-1/2*3):2
1/2*3*4=(1/1*2-1/2*3):2
...
cu nhu the cho den:
1/98*99*100=(1/98*99-1/99*100):2
suy ra : 1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100
=(1/1*2-1/2*3):2+(1/2*3-1/3*4):2+...+(1/98*99-1/99*100):2
=(1/1*2-1/2*3+1/2*3-1/3*4+...+1/98*99-1/99*100):2
=(1/1*2-1/99*100):2
=(1/2-1/9900)
=(4950/9000-1/9000):2
=4949/9000:2
=4949/18000
học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
Ta có:
\(A=2\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\right).\)
\(A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}.\)
\(A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}.\)
\(A=\left(\dfrac{1}{2.3}-\dfrac{1}{2.3}\right)+\left(\dfrac{1}{3.4}-\dfrac{1}{3.4}\right)+...+\left(\dfrac{1}{98.99}-\dfrac{1}{98.99}\right)+\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right).\)
\(A=0+0+...+0+\left(\dfrac{1}{1.2}-\dfrac{1}{99.100}\right).\)
\(A=\dfrac{1}{1.2}-\dfrac{1}{99.100}.\)
\(A=\dfrac{1}{2}-\dfrac{1}{9900}.\)
\(A=\dfrac{4950}{9900}-\dfrac{1}{9900}.\)
\(A=\dfrac{4949}{9900}.\)
Vậy \(A=\dfrac{4949}{9900}.\)
~ Chúc bn học tốt!!! ~
Bài mik đúng thì nhớ tick mik nha!!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt A là tên biểu thức
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(2A=\frac{1}{2}-\frac{1}{9900}\)
\(2A=\frac{4949}{9900}\)
\(A=\frac{4949}{9900}:2=\frac{4949}{19800}\)
úp tui giúp tui