Giá trị của A=1+4+42+43+....+42020 bằng:
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(4M=4^2+4^3+4^4+...+4^{2021}+4^{2022}\)
\(\Rightarrow4M-M=4^2+4^3+4^4+...+4^{2021}+4^{2022}-\left(4+4^2+4^3+...+4^{2020}+4^{2021}\right)\)
\(\Leftrightarrow3M=4^{2022}-4\)
\(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)
\(4M=4^2+4^3+...+4^{2022}\)
\(\Leftrightarrow M=\dfrac{4^{2022}-4}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
59 – 43 – 8 = 8
59 – 42 – 3 = 14
58 – 41 – 9 = 8
59 – 42 – 4 = 13
Đáp án cần chọn là A
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có M ⋮ 25 vì 75 ⋮ 25
Lại có M = 75 ( 42021 + 42020 + ... + 42 + 4 + 1 )
= 75 . 4 ( 22020 + 22019 + ... + 4 + 1 + 0,25 ) ⋮ 4 vì 4 ⋮ 4
Mà ( 25; 4 ) = 1 ⇒ M ⋮ 100
Vậy M ⋮ 100
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=75.4\left(4^{2020}+4^{2019}+...+4+1\right)+75+25=\)
\(=300.\left(4^{2020}+4^{2019}+...+4+1\right)+100=\)
\(=100\left[3.\left(4^{2020}+4^{2019}+...+4+1\right)+1\right]⋮100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, A = 2 10 - 2 5 = 1024 - 32 = 992
b, B = 4 3 - 4 2 - 4 = 64 - 16 - 4 = 44
c, C = 3 2 . 2 3 + 4 3 . 2 5 = 9.8 + 64.32 = 2120
d, D = 1 3 + 2 3 + 3 3 + 4 3 + 5 3 = 1 + 8 + 27 + 64 + 125 = 225
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = 2 10 - 2 5 = 1024 - 32 = 992 .
b) B = 4 3 - 4 2 - 4 = 64 - 16 - 4 = 44 .
c) C = 3 2 . 2 3 + 4 3 . 2 5 = 9 . 8 + 64 . 32 = 2120 .
d) D = 1 3 + 2 3 + 3 3 + 4 3 + 5 3 = 1 + 8 + 27 + 64 + 125 = 225
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) A = 2 10 - 2 5 = 1024 - 32 = 992 . b ) B = 4 3 - 4 2 - 4 = 64 - 16 - 4 = 44 . c ) C = 3 2 . 2 3 + 4 3 . 2 5 = 9 . 8 + 64 . 32 = 2120 . d ) D = 1 3 + 2 3 + 3 3 + 4 3 + 5 3 = 1 + 8 + 27 + 64 + 125 = 225 .
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
b) Gọi \(d\inƯC\left(21n+4;14n+3\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}21n+4⋮d\\14n+3⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}42n+8⋮d\\42n+9⋮d\end{matrix}\right.\)
\(\Leftrightarrow1⋮d\)
\(\Leftrightarrow d\inƯ\left(1\right)\)
\(\Leftrightarrow d\in\left\{1;-1\right\}\)
\(\LeftrightarrowƯCLN\left(21n+4;14n+3\right)=1\)
hay \(\dfrac{21n+4}{14n+3}\) là phân số tối giản(đpcm)
Bài 1:
a) Ta có: \(A=1+2-3-4+5+6-7-8+...-299-300+301+302\)
\(=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(297+298-299-300\right)+301+302\)
\(=\left(-4\right)+\left(-4\right)+...+\left(-4\right)+603\)
\(=75\cdot\left(-4\right)+603\)
\(=603-300=303\)
A = 1 + 4 + 4^2 + 4^3 + ... + 4^2020
4A = 4 + 4^2 + 4^3 + 4^4 + ... + 4^2020 + 4^2021
-
A = 1 + 4 + 4^2 + 4^3 + ... + 4^2020
=
3A = 4^2021 - 1
A = 4^2021 - 1 : 3
(Lưu ý : Dấu ^ là dấu mũ nhé)