Cho A=1.2+2.3+3.4+4.5+............+2017.2018 va B=2018 mu3/3 So sanh A va B
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A=1.2+2.3+3.4+4.5+...+2017.2018
=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3
3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018
3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)
=> 3A=2017.2018.2019 => \(A=\frac{2017.2018.2019}{3}\); \(B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)
Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018
=> 2017.2018.2019<2018.2018.2018
=> A<B
Bui The Hao lam dung roi
mk cung dang can bai nay
Thanks vi da dang honganh
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\(A=1.2+2.3+3.4+...+2017.2018\)
\(3A=1.2.3+2.3.3+3.4.3+...+2017.2018.3\)
\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+2017.2018.\left(2019-2016\right)\)
\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+2017.2018.2019-2016.2017.2018\)
\(\Rightarrow3A=2017.2018.2019\)
\(\Rightarrow A=\dfrac{2017.2018.2019}{3}\)
\(A=\dfrac{2017.2018.2019}{3};B=\dfrac{2018^3}{3}=\dfrac{2018.2018.2018}{3}\)
\(A=2739315938;B=2739316611\)
\(\Rightarrow A< B\)
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Ta có : A=1.2+2.3+3.4+....+2015.2016
=>3A= 1.2.3 + 2.3.3 + 3.4.3 + 4.5.3 + ... + 2017.2018.3
=>3A= 1.2.3 + 2.3.( 4 - 1 ) + 3.4.( 5-2 ) + 4.5.( 6-3 ) + ... 2017 . 2018 . ( 2019 - 2016 )
=>3A=-1.2.3 + 2.3.4 - 2.3.1 + 3.4.5 - 3.4.2 + 4.5.6 - 4.5.3 +.....+ 2017 . 2018 .2019 - 2017 . 2018 . 2016
=>A= 2017 . 2018 . 2019
![](https://rs.olm.vn/images/avt/0.png?1311)
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A = 1.2 + 2.3 + 3.4 + ....... + 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 = (1.2.3 - 1.2.3) + (2.3.4-2.3.4) + ... + (98.99.100 - 98.99.100) + 99 . 100 . 101
3A = 99 . 100 . 101 = 999900
A = 999900 : 3
A = 333300
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cho bài kham khảo nè :
A=1.2+2.3+3.4+4.5+...+2017.2018
=> 3A=1.2.3+2.3.3+3.4.3+4.5.3+...+2017.2018.3
3A=1.2.3+2.3(4-1)+3.4(5-2)+4.5(6-3)+...+2017.2018.(2019-2016)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+2017.2018.2019-2016.2017.2018
3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+2017.2018.2019)-(1.2.3+2.3.4+3.4.5+...+2016.2017.2018)
=> 3A=2017.2018.2019 => \(A=\frac{2017.2018.2019}{3};B=\frac{2018^3}{3}=\frac{2018.2018.2018}{3}\)
Ta có: 2017.2019=2017(2018-1)=2017.2018+2017<2017.2018+2018=2018(2017+1)=2018.2018
=> 2017.2018.2019<2018.2018.2018
=> A<B
thank nha
A=1.2+2.3+3.4+...+2017.2018
3A=1.2.3+2.3.3+3.4.3+...+2017.2018.3
3A=1.2.3+2.3.(4−1)+3.4.(5−2)+...+2017.2018.(2019−2016)
3A=1.2.3+2.3.4−1.2.3+3.4.5−2.3.4+...+2017.2018.2019−2016.2017.2018
⇒3A=2017.2018.2019
⇒A=2017.2018.20193
A=2017.2018.20193;B=201833=2018.2018.20183
A=2739315938;B=2739316611
⇒A<B