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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 3A = 1.2.3 + 2.3.3 + 3.4.3 + .... + n.( n + 1 ).3
=> 3A = 1.2.( 3 - 0 ) + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ..... + n.( n + 1 ).[ ( n + 2 ) - ( n - 1 ) ]
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ..... + n.( n + 1 ).( n + 2 ) - ( n - 1 ).n.( n + 1 )
=> 3A = ( 1.2.3 - 1.2.3 ) + ( 2.3.4 - 2.3.4 ) + .... + [ ( n - 1 ).n.( n + 1 ) - ( n - 1 ).n.( n + 1 ) ] + n.( n + 1 ).( n + 2 )
=> 3A = n.( n + 1 ).( n + 2 )
=> A = \(\frac{n.\left(n+1\right).\left(n+2\right)}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1.2+2.3+...+49.50
3A=1.2.3+2.3.3+...+49.50.3
3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)
3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48
3A=49.50.51
=>A=49.25.51
=>A=62475
A=1.2+2.3+...+49.50
3A=1.2.3+2.3.3+...+49.50.3
3A=1.2.(4-1)+2.3.(5-2)+....+49.50.(51-48)
3A=1.2.4-1.2.1+2.3.5-2.3.2+...+49.50.51-49.50.48
3A=49.50.51
=>A=49.25.51
=>A=62475
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+....+100.101.102-99.100.101
\(\Rightarrow\)3A=100.101.102
\(\Rightarrow\)A=343400
B=1.2+1+2.3+1+3.4+1+...+100.101+100
\(\Rightarrow\)B=1.2+2.3+3.4+...+100.101+[1+2+3+..+100]
Mặt khác 1.2+2.3+3.4+4.5+...+100.101=A
\(\Rightarrow\)B=343400+101.100:2=348450
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}=\frac{1}{1}-\frac{1}{50}=\frac{49}{50}\)
Vậy A=49/50
Công thức: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
S = 1.2 + 2.3 + 3.4 + ... + 99.100
3S = 1.2.3 + 2.3.3 + 3.4.3 + .. +99.100.3
3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
3S = 99.100.101
3S = 999 900
S = 333 300
Vậy S = 333 300
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5}{1.2}+\frac{13}{2.3}+\frac{25}{3.4}+\frac{41}{4.5}+...+\frac{181}{9.10}\) \(=\frac{4+1}{2}+\frac{12+1}{6}+\frac{24+1}{12}+\frac{40+1}{20}+...+\frac{180+1}{90}\)
\(=2+\frac{1}{1.2}+2+\frac{1}{2.3}+2+\frac{1}{3.4}+2+\frac{1}{4.5}+...+2+\frac{1}{9.10}\)
\(=18+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=19-\frac{1}{10}\)
\(=\frac{189}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\:\frac{-1}{1.2}+\frac{-1}{2.3}+\frac{-1}{3.4}+\frac{-1}{4.5}\)
\(=-1\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\right)\)
=\(-1\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)\)
=\(-1\left(1-\frac{1}{5}\right)\)
=\(-1\times\frac{4}{5}\)
=\(\frac{-4}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{50.51}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{50}-\frac{1}{51}\)
\(=\frac{1}{3}-\frac{1}{51}\)
\(=\frac{17}{51}-\frac{1}{51}\)
\(=\frac{16}{51}\)
Ta có: 3S = 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + .....+ 50.51.(52 -49)
= 1.2.3 - 0 + 2.3.4 - 1.2.3 + 3.4.5 -2.3.4 + .....+ 50.51.52 - 49.50.51
3S = 50.51.52
S = 50.17.52 =44200