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\(S=2^{2019}-2^{2018}-2^{2017}-...-2^2-2-1\)
\(=2^{2019}-\left(1+2+2^2+...+2^{2017}+2^{2018}\right)\) (1)
Đặt \(Q=1+2+2^2+...+2^{2017}+2^{2018}\)
\(2Q=2+2^2+2^3+...+2^{2018}+2^{2019}\)
\(2Q-Q=2^{2019}-1\)
\(Q=2^{2019}-1\)(2)
Từ (1) và (2), ta được:
\(S=2^{2019}-\left(2^{2019}-1\right)=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=2+2^2+...+2^{2016}\)
\(\Rightarrow2S=2.\left(2+2^2+...+2^{2016}\right)\)
\(\Rightarrow2S=2^2+2^3+...+2^{2017}\)
\(\Rightarrow2S-S=2^2+2^3+...+2^{2017}-\left(2+2^2+...+2^{2016}\right)\)
\(\Rightarrow S=2^2+2^3+...+2^{2017}-2-2^2-...-2^{2016}\)
\(\Rightarrow S=2^{2017-2}\)
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Bài 1a
B=4/1.3 + 4/3.5 + 4/5.7+...+4/2017.2019
B=4.2/(1.3).2 + 4.2/(3.5).2 + 4.2/(5.7).2+....+4.2/(2017.2019).2
B=2.( 2/1.3 + 2/3.5 + 2/5.7 +...+ 2/2017.2019 )
B=2.(1-1/3+1/3-1/5+1/5-1/7+....+1/2017-1/2019)
B=2.(1-1/2019)
B=2.(2019/2019-1/2019)
B=2.2018/2019
B=4036/2019
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\(2^0+2^1+2^2+2^3+2^0.2^1.2^2.2^3\)3
\(=1+2+4+8+1.2.4.8\)
\(=15+64=79\)
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n = 25.3x^2
Vì n có 60 ước nên ta có:
(5 + 1)(x2 + 1) = 60
=> 6(x2 + 1) = 60
=> x2 + 1 = 10
=> x2 = 9
=> x = 3 (Vì x thuộc N*)
Do đó n = 25.33^2 = 25.39 = 629856
S=1-2+2^2-2^3+...+2^2016
2S=2-22+23-24+...+22017
S+2S=(1-2+2^2-2^3+...+2^2016)+(2-22+23-24+...+22017)
3S=1-2+22-23+...+22016+2-22+23-24+...+22017
3S=1+22017
3S-22017=1+22017-22017=1