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\(A=1+2^1+2^2+...+2^{2017}\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(2A-A=2^{2018}-1hayA=2^{2018}-1\)
2; 3 tuong tu
1) A = 1 + 2 + 22 + 23 + .... + 22018
2A = 2 + 22 + 23 + 24 + ..... + 22019
2A - A = ( 2 + 22 + 23 + 24 + ..... + 22019 ) - ( 1 + 2 + 22 + 23 + .... + 22018 )
Vậy A = 22019 - 1
2) B = 1 + 3 + 32 + 33 + ..... + 32018
3A = 3 + 32 + 33 + ...... + 32019
3A - A = ( 3 + 32 + 33 + ...... + 32019 ) - ( 1 + 3 + 32 + 33 + ..... + 32018 )
2A = 32019 - 1
Vậy A = ( 32019 - 1 ) : 2
3) C = 1 + 4 + 42 + 43 + ...... + 42018
4A = 4 + 42 + 43 + ...... + 42019
4A - A = ( 4 + 42 + 43 + ...... + 42019 ) - ( 1 + 4 + 42 + 43 + ...... + 42018 )
3A = 42019 - 1
Vậy A = ( 42019 - 1 ) : 3
Đề sai ! Sửa \(\frac{1}{2}\)thành \(\frac{3}{2}\)
Bài giải
\(A=\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2018}\)
\(A=\frac{3}{2}+\frac{3^2}{2^2}+\frac{3^3}{2^3}+...+\frac{3^{2018}}{2^{2018}}\)
\(\frac{2}{3}A=1+\frac{3}{2}+\frac{3^2}{2^2}+...+\frac{3^{2017}}{2^{2017}}\)
\(A-\frac{2}{3}A=\frac{3^{2018}}{2^{2018}}-1\)
\(\frac{1}{3}A=\frac{3^{2018}}{2^{2018}}-1\)
\(A=\left(\frac{3^{2018}}{2^{2018}}-1\right)\cdot3=\frac{3^{2019}}{2^{2018}}-3\)
\(B=\left(\frac{3}{2}\right)^{2019}\text{ : }2=\frac{3^{2019}}{2^{2019}}\cdot\frac{1}{2}=\frac{3^{2019}}{2^{2020}}\)
\(B-A=\frac{3^{2019}}{2^{2020}}-\frac{3^{2019}}{2^{2018}}+3=3^{2019}\left(\frac{1}{2^{2018}}\cdot\frac{1}{2^4}-\frac{1}{2^{2018}}\right)+3=3^{2019}\left[\frac{1}{2^{2018}}\left(\frac{1}{2^4}-1\right)\right]+1\)
\(=3^{2019}\cdot\frac{1}{2^{2018}}\cdot\frac{-15}{16}+3\)
A=1+2+22+23+...+22018+22019
>2A=2(1+2+22+23+...+22018+22019)
=>2A=2+22+23+...+22018+22019
=>2A-A=(2+22+23+...+22019+22020)-(1 + 2 + 22 + 23 + ... + 22018 + 22019)
=>A=22020-1
B=1 + 32 + 34 + 36 +...+ 32018 + 32020
=>9B=3(1 + 32 + 34 + 36 +...+ 32018 + 32020)
=>9B=3+32 + 34 + 36 +...+ 32020 + 32022
=>9B-B=(3+32 + 34 + 36 +...+ 32018 + 32020)-(1 + 32 + 34 + 36 +...+ 32018 + 32020)
=.8B=32022-1
=>B=32022:8-1
\(A=1+3^1+3^2+...+3^{2017}\)
\(3A=3+3^2+3^3+...+3^{2018}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2018}\right)-\left(1+3^1+3^2+...+3^{2017}\right)\)
\(2A=3^{2018}-1\)
\(A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\)\(B-A=\frac{3^{2018}}{2}-\frac{3^{2018}-1}{2}=\frac{3^{2018}-3^{2018}+1}{2}=\frac{1}{2}\)
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
\(\Rightarrow A=\frac{3^{2018}-1}{2}\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{3^{2018-1}}{2}}{\frac{3^{2018}}{2}}=\frac{\frac{3^{2018}}{2}}{\frac{3^{2018}}{2}}-\frac{1}{\frac{3^{2018}}{2}}=1-\frac{1}{\frac{3^{2018}}{2}}\)
3A = 3+3^2+....+3^2018
2A=3A-A=(3+3^2+....+3^2018)-(1+3+3^2+....+3^2017) = 3^2018-1
=> A = (3^2018-1)/2
=> B-A = 3^2018-3^2018+1/2 = 1/2
Tk mk nha