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1. A = 2 + 22 + 23 + 24 +...+22019
2A= 2( 2 + 22 + 23 + 24 +...+22019)
2A= 22 + 23 + 24 +...+22019+22020
2A-A= (22 + 23 + 24 +...+22019+22020) - ( 2 + 22 + 23 + 24 +...+22019)
A= 22020-2
Vì 22020=22020 nên 22020-2 < 22020
=> A < B
Vậy..
Ta có:
\(2A=2^2+2^3+2^4+2^5+...+2^{2020}\)
\(\Leftrightarrow2A-A=\left(2^2+2^3+2^4+...+2^{2020}\right)-\left(2+2^2+2^3+....+2^{2019}\right)\)
\(\Leftrightarrow A=2^{2020}-2\)
\(\Rightarrow A< B\)
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\(A=1+3^1+3^2+...+3^{2017}\)
\(3A=3+3^2+3^3+...+3^{2018}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2018}\right)-\left(1+3^1+3^2+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\)\(B-A=\frac{3^{2018}}{2}-\frac{3^{2018}-1}{2}=\frac{3^{2018}-3^{2018}+1}{2}=\frac{1}{2}\)
Vậy \(B-A=\frac{1}{2}\)
Chúc bạn học tốt ~
ta có: A = 1 + 31 + 32 + ...+ 32017
=> 3A = 31 + 32 + 33 + ....+ 32018
=> 3A - A = 32018 - 1
\(\Rightarrow A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3^{2018-1}}{2}}{\frac{3^{2018}}{2}}=\frac{\frac{3^{2018}}{2}}{\frac{3^{2018}}{2}}-\frac{1}{\frac{3^{2018}}{2}}=1-\frac{1}{\frac{3^{2018}}{2}}\)
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\(\text{Đặt }A=\left(2^0+2^1+2^2+2^3+2^4\right)+...+\left(2^{5n-5}+2^{5n-4}+2^{5n-3}+2^{5n-2}+2^{5n-1}\right)\)
\(=\left(1+2 +4+8+16\right)+...+2^{5n-5}.\left(2^0+2^1+2^2+2^3+2^4\right)\)
\(=31+...+2^{5n-5}.31\)
\(=31.\left(1+...+2^{5n-5}\right)\text{chia hết cho 31}\left(đpcm\right)\)
tìm giá trị lớn nhất 1/x2 +2010
Xác định dấu của c để biểu thức 2a2bctrais dấu với biểu thức -3a5b3c2
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S = 2+2+22+23+............+299
2S = 22+22+23+............+2100
2S - S = S = 2100 = 25.295 = 32.295 > 10.295
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A = 2 + 22 + 23 + 24 + ... + 29 + 210
A = ( 2 + 22 ) + ( 23 + 24 ) + ... + ( 29 + 210 )
A = ( 1 + 2 ) . 2 + ( 1 + 2 ) . 23 + ... + ( 1 + 2 ) . 29
A = 3 . 2 + 3 . 23 + ... + 3 . 29
A = 3 . ( 2 + 23 + ... + 29 )
=> A chia hết cho 3
\(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+210\right)=2\left(2^0+2^1\right)+2^3\left(2^0+2^1\right)+... \)
\(2^0=1,2^1=2,2^0+2^1=3\)
chả hiểu gì vì mình mới học lớp 5