2/2.3+2/3.4+...+2/19.20 = ?
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(\Rightarrow A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)
\(\Rightarrow A=2.\left(1-\frac{1}{20}\right)\)
\(\Rightarrow A=2.\frac{19}{20}\)
\(\Rightarrow A=\frac{19}{10}\)
2.(1/1.2+1/2.3+.....+1/18.19+1/19.20)
2.(1/1-1/2+1/2-1/3+......+1/19-1/20)
2.(1/1-1/20)= 2.19/20=19/10
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{18.19}+\frac{2}{19.20}\)
\(A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(A=2\left(1-\frac{1}{20}\right)\)
\(A=2.\frac{19}{20}=\frac{19}{10}\)
Vậy ...
=2.(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+......+\(\frac{1}{19.20}\))
=2.( 1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+..........+\(\frac{1}{19}\)-\(\frac{1}{20}\))
=2.(1-\(\frac{1}{20}\))
=2.\(\frac{19}{20}\)
= \(\frac{19}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{1}-\frac{1}{20}\)
\(=\frac{19}{20}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(1.2+2.3+3.4+...+19.20\)
\(=\dfrac{20.\left(20+1\right).\left(20+2\right)}{3}\)
\(=3080\)
b) \(9+99+999+...+999...9\left(100so9\right)\)
\(\)\(=\left(10-1\right)+\left(100-1\right)+\left(1000-1\right)+...+\left(1000...0-1\right)\left(99so0\right)\)
\(=\left(10+10^2+10^3+...10^{99}\right)+\left(-1\right).100\)
\(=\left(1+10+10^2+10^3+...10^{99}\right)+\left(-1\right).101\)
\(=\dfrac{10^{99+1}-1}{99-1}-101\)
\(=\dfrac{10^{100}-1}{98}-101\)
\(=\dfrac{10^{100}-9899}{98}\)
c) \(999.9x222...2\) (100 số 9; 100 số 2)
\(9x2=18\)
\(99x22=2178\)
\(999x222=\text{221778}\)
\(9999x2222=22217778\)
\(99999x22222=2222177778\)
\(.........\)
Theo quy luật trên ta có 100 số 9 nhân 100 số 2:
\(999.9x222...2=222...21777...78\) (99 sô 2; 1 số 1; 99 số 7; 1 số 8)
![](https://rs.olm.vn/images/avt/0.png?1311)
D=1.2+2.3+3.4+...+19.20
=>3D=1.2.3+2.3.3+3.4.3+...+19.20
=1.2.3+2.3(4-1)+3.4(5-2)+...+19.20(21-18)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20
=19.20.21=7980
=>D=7980:3=2660
Vậy D=2660
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=1.2+2.3+3.4+...+19.20\)
Ta có: \(A=1.2+2.3+3.4+...+19.20\)
\(3A=1.2.3+2.3.3+3.4.3+...+19.20.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-1\right)+...+19.20.\left(21-1\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+19.20.21-18.19.20\)
\(3A=19.20.21\)
\(A=19.20.7\)
\(A=2660\)
\(1\cdot2+2\cdot3+...+19\cdot20=\frac{1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+...+19\cdot20\cdot\left(21-17\right)}{3}\)
\(=\frac{1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+...+19\cdot20\cdot21-18\cdot19\cdot20}{3}\)\(=\frac{19\cdot20\cdot21}{3}=2660\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3S=2.3.3+3.4.3+4.5.3+...+19.20.3\)
\(3S=2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+19.20.\left(21-18\right)\)
\(3S=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+19.20.21-18.19.20\)
\(3S=19.20.21-1.2.3\Rightarrow S=\frac{19.20.21-1.2.3}{3}=7.19.20-2\)