Tính tổng: 1/1.2.3+1/2.3.4|+...+1/98.99.100
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b) S = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : A = 1.2.3 + 2.3.4 + 4.5.6 + ..... + 98.99.100
=> 6A = 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + ...... + 98.99.100.101
=> 6A = 98.99.100.101
=> A = \(\frac{98.99.100.101}{6}=16331700\)
có 20172 đồng dư 1 mod (3)
=> (20172)50 đồng dư 1 mod (3)
=> (20172)50-1 đồng dư 1-1 = 0 mod (3)
=> dpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt N=1/1.2.3+1/2.3.4+....+1/98.99.100
=1/2.(2/1.2.3+2/2.3.4+...+2/98.99.100)
=1/2(1/1.2-1/2.3+1/3.4+...+1/98.99-1/99.100)
=1/2(1/2-1/99.100)
=1/2.4949/9900
=4949/19800
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B=1/1.2.3+1/2.3.4+1/3.4.5+............+1/98.99.100
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(=\frac{1}{2}\cdot\frac{4949}{9900}\)
\(=\frac{4949}{19800}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(B=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
\(B=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(B=\frac{1}{2}.\frac{4949}{9900}=\frac{4949}{19800}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta xét:
\(\frac{1}{1.2}-\frac{1}{2.3}=\frac{2}{1.2.3};\frac{1}{2.3}-\frac{1}{3.4}=\frac{2}{2.3.4};...;\frac{1}{98.99}-\frac{1}{99.100}=\frac{2}{98.99.100}\)
Qua công thức trên, bạn có thể rút ra tổng quát: (đây là mình nói thêm)
\(\frac{1}{n.\left(n+1\right)}-\frac{1}{\left(n+1\right).\left(n-2\right)}=\frac{2}{n.\left(n+1\right).\left(n+2\right)}\)
Ta suy ra:
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
Thấy \(-\frac{1}{2.3}+\frac{1}{2.3}=0;-\frac{1}{3.4}+\frac{1}{3.4}=0;...\)
\(\Rightarrow2B=\frac{1}{2}-\frac{1}{99.100}=\frac{1}{2}-\frac{1}{9900}=\frac{4950}{9900}-\frac{1}{9900}=\frac{4949}{9900}\)
\(\Rightarrow B=\frac{4949}{9900}:2=\frac{4949}{19800}\)
Mình nhầm, công thức tổng quát mình nói thêm bạn đổi cái n-2 thành n+2 nha
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\)
\(S=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)
\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)
\(S=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)
\(\Rightarrow2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4849}{9900}\)
\(\Rightarrow S=\frac{4949}{9900}\div2=\frac{4949}{19800}\)
nhân tổng trên cho 2 ta có;
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
=1/1.2-1/99.100
=4949/9900
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