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A=\((1+2)+\left(2^2+2^3\right)+...+\left(2^{19}+2^{20}\right)\)
A=\(3.1+2^2\left(1+2\right)+...+2^{19}\left(1+2\right)\)
A=\(3.1+3.2^2+...+3.2^{19}\)
A=\(3\left(1+2^2+...+2^{19}\right)\)\(⋮3\)
Vậy A\(⋮3\)
A=(1+2)+(22+23)+...+(219+220)(1+2)+(22+23)+...+(219+220)
A=3.1+22(1+2)+...+219(1+2)3.1+22(1+2)+...+219(1+2)
A=3.1+3.22+...+3.2193.1+3.22+...+3.219
A=3(1+22+...+219)3(1+22+...+219)⋮3⋮3
NÊN A⋮3
Câu 9: Năm nay tổng số tuổi của hai ông cháu là 85 tuổi. Ông hơn cháu 61 tuổi. Hiên nay tuổi ông là:
Giải giúp mình với plaeeeeeeeeeeeeeeeeeeeee
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}\)
\(=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{99-98}{98.99}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
\(=1-\dfrac{1}{99}\)
\(A=\dfrac{2021}{2022}=\dfrac{2022-1}{2022}=1-\dfrac{1}{2022}\)
Có \(2022>99>0\Leftrightarrow\dfrac{1}{99}>\dfrac{1}{2022}\)
Suy ra \(A>B\).
Sửa đề: \(A=2+2^2+2^3+2^4+...+2^{19}+2^{20}\)
=>\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{19}+2^{20}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{19}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{19}\right)⋮3\)
1, Ta có :
A = 1.2+2.3+3.4+4.5+...+100.101
B = 1.3+2.4+3.5+4.6+....+100.102
=> B - A = ( 1.2+2.3+3.4+4.5+...+100.101) - (1.3+2.4+3.5+4.6+...+100.102)
=> B - A = 1.2+2.3+3.4+4.5+...+100.101-1.3-2.4-3.5-4.6-....-100.102
=> B - A = 1.2+(2.3-1.3)+(3.4-2.4)+(4.5-3.5)+...+(100.101-99.101)-100.102
=> B - A = 2+3+4+5+...+101-10200
=> B - A = (2+101)+(3+100)+...+(51+52)-10200
=> B - A = 103+103+103+....+103-10200 ( 50 SỐ 103 )
=> B - A = 103.50-10200
=> B - A = 5150-10200
=> B - A = -5050
A = 1.2 + 2.3 + 3.4 + ... + 100.101
3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 100.101.3
3A = 1.2.3 + 2.3(4 - 1) + 3.4.(5 - 2) + ... + 100.101.(102 - 99)
3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 100.101.102 - 99.100.101
3A = 100.101.102
A = 34340
B = 1.2.3 + 2.3.4 + 3.4.5 + ... + 100.101.102
4B = 1.2.3.4 + 2.3.4.4 + 3.4.5.4 + ... + 100.101.102.4
4B = 1.2.3.4 + 2.3.4(5 - 1) + 3.4.5.(6-2) + ... + 100.101.102.(103 - 99)
4B = 1.2.3.4 + 2.3.4.5 - 1.2.3.4 + 3.4.5.6 - 2.3.4.5 + ... + 100.101.102.103 - 99.100.101.102
4B = 100.101.102.103
B = 26527650
B/A
\(=\dfrac{1+\dfrac{2020}{2}+1+\dfrac{2019}{3}+...+1+\dfrac{1}{2021}+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}\)
\(=\dfrac{2022\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2021}+\dfrac{1}{2022}}=2022\)
a/
2020.2021=(2019+1)(2022-1)=
=2019.2022-2019+2022-1=2019.2022+2>2019.2022
b/
\(4^7=\left(2^2\right)^7=2^{14}< 2^{15}\)
c/
\(199^{20}< 200^{20}=\left(8.25\right)^{20}=\left(2^3.5^2\right)^{20}=2^{60}.5^{40}\)
\(2000^{15}=\left(16.125\right)^{15}=\left(2^4.5^3\right)^{15}=2^{60}.5^{45}\)
\(\Rightarrow2000^{15}=2^{60}.5^{45}>2^{60}.5^{40}>199^{20}\)
d/
\(31^{31}< 32^{31}=\left(2^5\right)^{31}=2^{155}\)
\(17^{39}>16^{39}=\left(2^4\right)^{39}=2^{156}\)
\(\Rightarrow17^{39}=2^{156}>2^{155}>31^{31}\)
A = 2 + 22 + 23 + 24 + ... + 219 + 220
A = (2 + 22) + (23 + 24) +... + (219 + 220)
A = 2.(1+2) + 23.(1 + 2) +... + 219.(l + 2)
A = 2.3 + 23.3 +...+ 219.3 Do đó A chia hết cho 3
A=1+2+3+...+2020+2021
A=(1+2021)[(2021-1):1+1]:2
A=2043231