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![](https://rs.olm.vn/images/avt/0.png?1311)
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`S = 1.2 + 2.3 + 3.4 + 4.5 + ... + 99.100.`
`3S = 1.2.3 + 2.3.(4-1) + 3.4.(5-4) + 4.5.(6-3) + ... + 99.100.(101-98)`
`3S = 1.2.3 + 2.3.4-1.2.3 + 3.4.5-4.5.6 + 4.5.6-3.4.5 + ... + 99.100.101-98.99.100`
`3S = 99.100.101`
`S = 33.100.101`
`S = 333300`
3S=1.2(3-0)+2.3(4-1)+.....+99.100(101-98)
=1.2.3-0.1.2+2.3.4-1.2.3+4.5.6-2.3.4+....+99.100.101-98-99-100
=99.100.101
S=33.100.101
=333300
![](https://rs.olm.vn/images/avt/0.png?1311)
S=1.2+ 2.3+4,5.......+99.100
Nhân cả 2 vế với 3, ta được:
3S=1.2.3+ 2.3.3+ 3.4.3+ 4.5.3+...... 99.100.3
= 1.2.3 + 2.3(4-1) + 3.4.(5-2) +...+ 99.100.(101-98)
= 1.2.3 + 2.3.4 -1.2.3 + 3.4.5-2.3.4 +...+ 99.100.101-98.99.100
= 99.100.101
----> S = (99.100.101):3
S= 333300
Vậy A=333300
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3A=1.2.3+2.3.3+3.4.3+4.5.3+.....+9.10.3
3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+.....+9.10.(11-8)
3A=1.2.3-1.2.0+2.3.4-1.2.3+.....+9.10.11-9.10.8
3A=9.10.11
A=(9.10.11):3
A=330
CHẮC CHẮN 100% LÀ ĐÚNG
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{Ta có: A = 1.2+2.3+3.4+4.5+...+99.100 }\)
=> 3A = 3.(1.2+2.3+3.4+4.5+...+99.100)
=> 3A = 1.2.(3 - 0) +2.3.(4 - 1) + 3.4.(5-2) + ........ + 99.100.(101 - 98)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + .......... + 99.100.101
=> 3A = 99.100.101
\(\Rightarrow A=\frac{99.100.101}{3}=333300\)
k mình nếu đúng OK
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 1.2 + 2.3 + 3.4 + ... + 2008.2009
= ( 1.2.3 + 2.3.3 + 3.4.3 + ... + 2008.2009.3 ) :3
= [ 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5 - 2 ) + ... + 2008.2009.( 2010 - 2007 )] : 3
= [ 1.2.3 + 2.3.4 - 2.3.1 + 2.4.5 - 3.4.2 + ... + 2008.2009.2010 - 2008.2009.2007 ] : 3
= ( 2008.2009.2010 ) :3
= 2702828240
A= 1.2 +2.3+ 3.4 +4.5+...+ 99.100
3A = 1.2.3 +2.3.3+ 3.4.3+...+99.100.3
3A = 1.2.3 +2.3(4-1)+ 3.4(5-2)+...+99.100(101-98)
3A = 1.2.3 +2.3.4-1.2.3+ 3.4.5-2.3.4+...+99.100.101-98.99.100
3A = 999900
A = 333300
A= 1. 2 + 2.3 +3.4 + 4.5 + 99.100
A=2 +6+12+20 + 9900
A = 9930
bn nha
tk mk nha