Cho A=1+1/2+1/3 +1/4+...+1/2^2018 - 1 . Chứng minh rằng:: A < 2018
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Dễ thấy A > 1
Ta có:
\(A=\frac{1}{1^2}+\frac{1}{2^3}+...+\frac{1}{2018^{2019}}\)
\(< \frac{1}{1^2}+\frac{1}{2^2}+...+\frac{1}{2018^2}< 1+\frac{1}{1\cdot2}+...+\frac{1}{2017\cdot2018}\)
\(=1+1-\frac{1}{2}+...+\frac{1}{2018}=2-\frac{1}{2018}< 2\)
Vì \(1< A< 2\) nên A không nguyên
![](https://rs.olm.vn/images/avt/0.png?1311)
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2018 A = 2018 - 2018^2 + 2018^3 +...- 2018^2018 + 2018^2019
=> A + 2018 A = 1 +2018^2019
=> 2019 A = 1 + 2018^2019
=> 2019 A - 1 = 2018^2019
=> 2019 A -1 là 1 lũy thừa của 2018
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+2^1+2^2+...+2^{2017}\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2^{2018}-1hayA=2^{2018}-1\)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
mình nhầm \(A=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2^{2018}-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+2+2^2+2^3+...+2^{2017}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2018}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2018}\right)-\left(1+2+2^2+...+2^{2017}\right)\)
\(\Rightarrow A=2^{2018}-1\left(đpcm\right)\)
\(A=1+2+2^2+2^3+...+2^{2017}\)
\(\Rightarrow2A=2+2^2+2^3+2^4+...+2^{2018}\)
\(\Rightarrow2A-A=\left(2+2^2+2^3+...+2^{2018}\right)\)\(-\left(1+2+2^2+...+2^{2017}\right)\)
\(\Rightarrow A=2^{2018}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}=\left(1+\frac{1}{2018}\right)+\left(\frac{1}{2}+\frac{1}{2017}\right)+...+\left(\frac{1}{1009}+\frac{1}{1010}\right)\)
\(=\frac{2019}{1.2018}+\frac{2019}{2.2017}+...+\frac{2019}{1009.1010}\)
\(=2019\left(\frac{1}{1.2018}+\frac{1}{2.2017}+...+\frac{1}{1009.1010}\right)\)
Do đó \(A=1.2.3....2018.2019\left(\frac{1}{1.2018}+\frac{1}{2.2017}+...+\frac{1}{1009.1010}\right)⋮2019\) (đpcm)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^{2018}-1}\)
\(A=1+\left(\frac{1}{2}+\frac{1}{3}\right)+...+\left(\frac{1}{2^k}+\frac{1}{2^k+1}+...+\frac{1}{2^{k+1}-1}\right)+...+\left(\frac{1}{2^{2017}}+...+\frac{1}{2^{2018}-1}\right)\)
\(A< 1+\left(\frac{1}{2}+\frac{1}{2}\right)+...+\left(\frac{1}{2^k}+\frac{1}{2^k}+...+\frac{1}{2^k}\right)+...+\left(\frac{1}{2^{2017}}+...+\frac{1}{2^{2017}}\right)\)
\(=1+1+...+1=2018\).