S=1+2+22+23+...+22021
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VN
1
27 tháng 8 2023
\(S=1+2+2^2+2^3+2^4+...+2^{2011}\)
\(\Rightarrow S=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+...+2^{2009}\left(1+2+2^2\right)\)
\(\Rightarrow S=7+2^3.7+...+2^{2009}.7\)
\(\Rightarrow S=7\left(1+2^3+...+2^{2009}\right)⋮7\)
\(\Rightarrow dpcm\)
PP
7 tháng 5 2021
2A=2*(1+2+22+...+22020)=2+22+...+22021
2A-A=(1+2+22+...+22021)-(1+2+22+...+22020)
A=22021-1<2021
8 tháng 5 2021
Giải:
A=1+2+22+23+...+22020
2A=2+22+23+24+...+22021
2A-A=(2+22+23+24+...+22021)-(1+2+22+23+...+22020)
A=22021-1
⇒A<22021
Chúc bạn học tốt!
PK
1
11 tháng 5 2023
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\)
\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}.\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\)
\(\Rightarrow A-\dfrac{1}{2}A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2020}}+\dfrac{1}{2^{2021}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{2021}}+\dfrac{1}{2^{2022}}\right)\)\(\Rightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{2022}}\)
\(\Rightarrow\dfrac{1}{2}A=\dfrac{2^{2021}-1}{2^{2022}}\)
\(\Rightarrow A=\dfrac{2^{2021}-1}{2^{2023}}.2=\dfrac{2^{2021}-1}{2^{2021}}\)
Vậy \(A=\dfrac{2^{2021}-1}{2^{2021}}\)
NT
2
18 tháng 4 2022
A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60
22 tháng 3 2024
Ta có: A=12+122+123+124+...+122021+122022�=12+122+123+124+...+122021+122022
⇒2A=1+12+122+123+...+122020+122021⇒2�=1+12+122+123+...+122020+122021
⇒2A−A=(1+12+122+123+...+122020+122021)−(12+122+123+124+...+122021+122022)⇒2�-�=(1+12+122+123+...+122020+122021)-(12+122+123+124+...+122021+122022)
⇒A=1−122022<1⇒�=1-122022<1
⇒A<1 (1)⇒�<1 (1)
Lại có: B=13+14+15+1760�=13+14+15+1760
⇒B=1615⇒�=1615
⇒B=1+115>1⇒�=1+115>1
⇒B>1 (2)⇒�>1 (2)
Từ (1)(1) và (2)⇒A<B(2)⇒�<�
Vậy A<B
TM
1
5 tháng 11 2021
\(2P=2+2^2+2^3+...+2^{2022}\)
\(\Leftrightarrow P=2^{2022}-1< Q\)
TM
1
5 tháng 11 2021
\(2P=2+2^2+2^3+...+2^{2022}\)
\(\Leftrightarrow P< Q\)
VV
3
T
19 tháng 10 2023
\(A=2+2^2+2^3+...+2^{2020}+2^{2021}+2^{2022}\\=(2+2^2)+(2^3+2^4)+(2^5+2^6)+...+(2^{2021}+2^{2022})\\=2\cdot(1+2)+2^3\cdot(1+2)+2^5\cdot(1+2)+...+2^{2021}\cdot(1+2)\\=2\cdot3+2^3\cdot3+2^5\cdot3+...+2^{2021}\cdot3\\=3\cdot(2+2^3+2^5+..+2^{2021})\)
Vì \(3\cdot\left(2+2^3+2^5+...+2^{2021}\right)⋮3\)
nên \(A⋮3\).
\(Toru\)
TN
19 tháng 10 2023
A=(2+22)+22(2+22)+...+22020(2+22)
A= 6.1+22.6+...+22020.6
A=6(1+22+...+22020) chia hết cho 3
vậy A chia hết cho 3
\(2S=2+2^2+...+2^{2022}\\ \Leftrightarrow2S-S=S=2^{2022}-1\)