A = 2 + 22 + 23 + ... 22019
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=1+2+2^2+...+2^{2018}\)
\(2A=2+2^3+2^4+...+2^{2019}\)
\(A=2A-A=1-2^{2019}\)
\(B-A=2^{2019}-\left(1-2^{2019}\right)\)
\(B-A=2^{2019}-1+2^{2019}\)
\(B-A=1\)
`#3107`
\(A=1+2+2^2+2^3+...+2^{2018}\) và \(B=2^{2019}\)
Ta có:
\(A=1+2+2^2+2^3+...+2^{2018}\)
\(2A=2+2^2+2^3+...+2^{2019}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2019}\right)-\left(1+2+2^2+2^3+...+2^{2018}\right)\)
\(A=2+2^2+2^3+...+2^{2019}-1-2-2^2-2^3-...-2^{2018}\)
\(A=2^{2019}-1\)
Vậy, \(A=2^{2019}-1\)
Ta có:
\(B-A=2^{2019}-2^{2019}+1=1\)
Vậy, `B - A = 1.`
![](https://rs.olm.vn/images/avt/0.png?1311)
=> 2A =2 + 22 + 23 + ... + 22020
=> 2A-A =( 2 + 22 + 23 + ... + 22020)- (1 + 2 + 22 + 23 + ... + 22019)
=> A =22020-1
=> A+1 =22020
Vậy A + 1 là một số chính phương
![](https://rs.olm.vn/images/avt/0.png?1311)
Chọn D
Ta có u n là cấp số nhân công bội bằng 2.
Ta có
2 4 u 1 + 1 + 2 3 - 2 u 2 = 2 4 u 1 + 1 + 2 3 - 4 u 1 ≥ 2 2 4 u 1 + 1 . 2 3 - 4 u 1 = 8
Mặt khác
Suy ra
Vậy 2 4 u 1 + 1 + 2 3 - 2 u 2 = 8 log 2 ( 2 u 3 2 - 8 u 2 + 4 )
suy ra giá trị nhỏ nhất của n cần tìm là 2021
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 23 + (-77) + (-23) + 77 =
[23 + (-23)] + [(-77) + 77]
= …0+0=0……
b) (-2 020) + 2 021 + 21 + (-22)
=[(-2 020) + 2 021] + [21 + (-22)]
= …1……+ (-1)……..
= 0.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2VT=2^{x+1}+2^{x+2}+2^{x+3}+...+...+2^{x+2016}\)
\(VT=2VT-VT=2^{x+2016}-2^x=2^{2016}.2^x+2^x=2^x\left(2^{2016}+1\right)\)
\(VP=2^{2019}-2^3=2^3\left(2^{2016}-1\right)\)
\(\Rightarrow2^2\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)
\(\Rightarrow2^x=2^3\Rightarrow x=3\)
\(2^x+2^{x+1}+2^{x+2}+2^{x+2015}=2^{2019}-8\left(1\right)\)
Đặt \(S=2^x+2^{x+1}+2^{x+2}+2^{x+2015}\)
\(\Rightarrow S+\left(1+2^2+...2^{x-1}\right)=\left(1+2^2+...2^{x-1}\right)+2^x+2^{x+1}+2^{x+2}+2^{x+2015}\)
\(\Rightarrow S+\dfrac{2^{x-1+1}-1}{2-1}=1+2^2+...2^{x-1}+2^x+2^{x+1}+2^{x+2}+2^{x+2015}\)
\(\Rightarrow S+2^x-1=\dfrac{2^{x+2015+1}-1}{2-1}\)
\(\Rightarrow S+2^x-1=2^{x+2016}-1\)
\(\Rightarrow S=2^{x+2016}-2^x\)
\(\left(1\right)\Rightarrow2^{x+2016}-2^x=2^{2019}-8=2^{2019}-2^3\)
\(\Rightarrow2^x\left(2^{2016}-1\right)=2^3\left(2^{2016}-1\right)\)
\(\Rightarrow2^x=2^3\Rightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{21}{22}+\dfrac{22}{23}=\dfrac{967}{506}>1\)
\(B=\dfrac{21+22}{22+23}=\dfrac{43}{45}< 1\)
Vậy \(A>B\)
\(\dfrac{21}{22}\) > \(\dfrac{21}{22+23}\)
\(\dfrac{22}{23}\) > \(\dfrac{22}{22+23}\)
Cộng vế với vế ta có:
A = \(\dfrac{21}{22}\) + \(\dfrac{22}{23}\) > \(\dfrac{21+22}{22+23}\) = B ⇒ A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3.5^2+15.2^2-26\div2\)
= 3.25 + 15.4 - 13
= 75 + 60 - 13
= 135 - 13
= 122
b) \(5^3.2-100\div4+2^3.5\)
= 125.2 - 25 + 8.5
= 250 - 25 + 40
= 225 + 40
= 265
c)\(6^2\div9+50.2-3^3.33\)
= 36 : 9 + 100 - 9.33
= 4 + 100 - 297
= 104 - 297
= -193
d)\(3^2.5+2^3.10-81\div3\)
= 9.5 + 8.10 - 27
= 45 + 80 - 27
= 125 - 27
= 98
e) \(5^{13}\div5^{10}-25.2^2\)
= 53 - 25.4
= 125 - 100
= 25
f) \(20\div2^2+5^9\div5^8\)
= 20 : 4 + 5
= 5 + 5
= 10
\(A=2+2^2+2^3+...+2^{2019}\)
\(2A=2^2+2^3+...+2^{2020}\)
\(2A-A=2^2+2^3+...+2^{2020}\)\(-2-2^2-2^3-...-2^{2019}\)
\(A=2^{2020}-2\)
Thank you