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\(A=1+2^1+2^2+...+2^{2017}\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2^{2018}-1hayA=2^{2018}-1\)
2; 3 tuong tu
1) A = 1 + 2 + 22 + 23 + .... + 22018
2A = 2 + 22 + 23 + 24 + ..... + 22019
2A - A = ( 2 + 22 + 23 + 24 + ..... + 22019 ) - ( 1 + 2 + 22 + 23 + .... + 22018 )
Vậy A = 22019 - 1
2) B = 1 + 3 + 32 + 33 + ..... + 32018
3A = 3 + 32 + 33 + ...... + 32019
3A - A = ( 3 + 32 + 33 + ...... + 32019 ) - ( 1 + 3 + 32 + 33 + ..... + 32018 )
2A = 32019 - 1
Vậy A = ( 32019 - 1 ) : 2
3) C = 1 + 4 + 42 + 43 + ...... + 42018
4A = 4 + 42 + 43 + ...... + 42019
4A - A = ( 4 + 42 + 43 + ...... + 42019 ) - ( 1 + 4 + 42 + 43 + ...... + 42018 )
3A = 42019 - 1
Vậy A = ( 42019 - 1 ) : 3
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1x2x3x...2018x2019 - 1x2x3x..2018 - 1x2x3x4x...x2017x20182
= 1x2x3x...x2018x(2019 - 1 - 2018)
= 1x2x3x...x2018x0
= 0
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đặt A=1+4+4^2+4^3+...+4^2018
B=1+2+2^2+2^3+...+2^2018
A=1+4+4^2+4^3+...+4^2018
4A=4+4^2+4^3+...+4^2019
4A-A=(4+4^2+4^3+...+4^2019)-(1+4+4^2+4^3+...+4^2018)
3A=4^2019-1
A=(4^2019)/3
B=1+2+2^2+2^3+...+2^2018
2B=2+2^2+2^3+...+2^2019
2B-B=(2+2^2+2^3+...+2^2019)-(1+2+2^2+2^3+...+2^2018)
B=2^2019-1
=>(1+4+4^2+4^3+...+4^2018)/(1+2+2^2+2^3+...+2^2018) =A/B=(4^2019-1)/3/(2^2019-1)
=(4^2019-1)/(3.2^2019-3)
Vậy ...............................
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+) Gọi A là tổng của dãy số: 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018.
+) Số số hạng của A là:
A = (2018 - 1) : 1 + 1 = 2018.
+) Tổng A là: (2018 + 1). 2018 : 1 = 4074342.
Vậy, A = 4074342 (hay 1+ 2 + 3 + 4 + ... + 2016 + 2017 + 2018 = 4074342).
![](https://rs.olm.vn/images/avt/0.png?1311)
cái này là khi chiều mới thi nầy
Giải:
Ta có:A=1.2+2.3+3.4+...+2017.2018
3A=1.2.3 2.3.3+...+2017.2018.3
=1.2.(3-0)+2.3.(4-1)+...+2017.2018.(2019-2016)
=1.2.3+2.3.4+...+2017.2018.2019-1.2.0-2.3.1-...-2017.2018.1016
=2017.2018.2019-1.2.0
=2017.2018.2019
=>A=2017.2018.2019/3=2018.(2017.2019)/3
Và B=20183/3=2018.2018.2018/3=2018.(2018.2018)/3
Lại có: 2017.2019=2017.(2018+1)=2017.2018+2017
2018.2018=(2017+1).2018=2017.2018+2018
Mà 2017.2018+2017<2017.2018+2018 =>2017.2019<2018.2018
=>2018.(2017.2019)<2018.(2018.2018)
=>A=2018.(2017.2019)/3<2018.(2018.2018)/3=B
=>A<B
Vậy A<B
Chúc Công Chúa Bloom học giỏi!!!
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\(A=\frac{1}{2018}+\frac{2}{2017}+...+\frac{2017}{2}+2018\)
\(=\left(\frac{1}{2018}+1\right)+\left(1+\frac{2}{2017}\right)+...+\left(\frac{2017}{2}+1\right)+1\)(2018 số hạng 1)
\(=\frac{2019}{2018}+\frac{2019}{2017}+...+\frac{2019}{2}+\frac{2019}{2019}=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)\)
Mà \(B=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\)
=> Khi đó : \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
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