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Lời giải:
$A-1=4+4^2+4^3+...+4^{2020}+4^{2021}$
$4(A-1)=4^2+4^3+4^4+....+4^{2021}+4^{2022}$
$\Rightarrow 4(A-1)-(A-1)=4^{2022}-4$
$3(A-1)=4^{2022}-4$
$\Rightarrow 3A+1=4^{2022}\vdots 4^{2021}$
a) Ta có A = 1 + 21 + 22 + ... + 22021
2A = 21 + 22 + 23 + ... + 22022
Vậy 2A = 21 + 22 + 23 + ... + 22022
b) 2A - A = ( 21 + 22 + 23 + ... + 22022 ) - ( 1 + 21 + 22 + ... + 22021 )
A = 22022 - 1
Vậy A = 22022 - 1
a)
\(A=1+2^1+2^2+2^3+...+2^{2020}+2^{2021}\)
\(2A=2^1+2^2+2^3+2^4+...+2^{2021}+2^{2022}\)
b)
\(2A=2^1+2^2+2^3+...+2^{2022}\)
\(2A-A=\left(2^1+2^2+2^3+...+2^{2022}\right)-\left(1+2^1+2^2+....+2^{2021}\right)\)
\(A=2^{2022}-1\)
=> đpcm
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
ta có :
\(1-\frac{2}{2.3}=\frac{2.3-2}{2.3}=\frac{1.2}{2.3}\)
tương tự : \(1-\frac{2}{3.4}=\frac{2.3}{3.4},....,1-\frac{2}{2020.2021}=\frac{2019.2020}{2020.2021}\)
Vậy \(S=\frac{1.2}{2.3}.\frac{2.3}{3.4}.....\frac{2019.2020}{2020.2021}=\frac{1.\left(2.3...2019\right)^2.2020}{2.\left(3.4....2020\right)^2.2021}=\frac{2}{2020.2021}\)
Chứng minh rằng: A = 3^2 + 3^3 + 3^4 + 3^5 + … + 3^2020 + 3^2021 chia hết cho 36 - Hoc24
\(A=\left(3^2+3^3\right)+3^2\left(3^2+3^3\right)+...+3^{2018}\left(3^2+3^3\right)\)
\(=36+3^2.36+...+3^{2018}.36=36\left(1+3^2+...+3^{2018}\right)⋮36\)
\(A=\left(3^2+3^3\right)+\left(3^4+3^5\right)+...+\left(3^{2020}+3^{2021}\right)\\ A=\left(3^2+3^3\right)+3^2\left(3^2+3^3\right)+...+3^{2018}\left(3^2+3^3\right)\\ A=\left(3^2+3^3\right)\left(1+3^2+...+3^{2018}\right)\\ A=36\left(1+3^2+...+3^{2018}\right)⋮36\)
Chọn B
Cảm ơn nha