A= 1+3+32+33+...+399
Chứng minh A chi hết cho 40
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16 tháng 10 2021
a: \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3+3^2+...+3^{119}\right)⋮3\)
b: \(B=3+3^2+3^3+3^4+...+3^{2020}\)
\(=3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)
\(=4\cdot\left(3+...+3^{2019}\right)⋮4\)
16 tháng 11 2021
\(1,Y=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{96}+3^{97}+3^{98}\right)\\ Y=\left(1+3+3^2\right)\left(1+3^3+...+3^{96}\right)\\ Y=13\left(1+3^3+...+3^{96}\right)⋮13\\ 2,A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{2018}+3^{2019}\right)\\ A=\left(1+3\right)\left(1+3^2+...+3^{2019}\right)\\ A=4\left(1+3^2+...+3^{2019}\right)⋮4\\ 3,\Leftrightarrow2\left(x+4\right)=60\Leftrightarrow x+4=30\Leftrightarrow x=36\)
B
cho A = 1 + 3 + 32 + 33 + ... + 311
a ) chứng minh A chia hết cho 13
b) chứng minh A chia hết cho 40
3
23 tháng 10 2023
A=1+3+3^2+3^3+...+3^98+3^99+3^100
A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)
A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)
A=13x3^3x13+...+3^98x13
=> 13x(1+3+3^3+...+3^98)chia hết cho 13
Vậy A chia hết cho 13
KV
23 tháng 12 2023
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
8 tháng 11 2023
`#3107.101107`
\(A=1+3+3^2+3^3+...+3^{101}\)
$A = (1 + 3 + 3^2) + (3^3 + 3^4 + 3^5) + ... + (3^{99} + 3^{100} + 3^{101}$
$A = (1 + 3 + 3^2) + 3^3 (1 + 3 + 3^2) + ... + 3^{99}(1 + 3 + 3^2)$
$A = (1 + 3 + 3^2)(1 + 3^3 + ... + 3^{99})$
$A = 13(1 + 3^3 + ... + 3^{99})$
Vì `13(1 + 3^3 + ... + 3^{99}) \vdots 13`
`\Rightarrow A \vdots 13`
Vậy, `A \vdots 13.`
T
8 tháng 11 2023
\(A=1+3+3^2+3^3+3^4+3^5+...+3^{101}\\=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+...+(3^{99}+3^{100}+3^{101})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+...+3^{99}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+...+3^{99}\cdot13\\=13\cdot(1+3^3+3^6+...+3^{99})\)
Vì \(13\cdot(1+3^3+3^6...+3^{99}\vdots13\)
nên \(A\vdots13\)
\(\text{#}Toru\)
HT
1
16 tháng 12 2021
\(A=1+3+3^2+3^3+...+3^{102}+3^{103}\)
\(\Rightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{102}+3^{103}\right)\)
\(\Rightarrow A=\left(1+3\right)+3^2\left(1+3\right)+...+3^{102}\left(1+3\right)\)
\(\Rightarrow A=\left(1+3\right)\left(1+3^2+...+3^{102}\right)\)
\(\Rightarrow A=4\left(1+3^2+...+3^{102}\right)⋮4\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
1 tháng 8 2023
Bài 1:
\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)
Bài 2:
\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)
1 tháng 8 2023
Bài 1 :
\(2^{49}=\left(2^7\right)^7=128^7\)
\(5^{21}=\left(5^3\right)^7=125^7\)
mà \(125^7< 128^7\)
\(\Rightarrow2^{49}>5^{21}\)
Bài 2 :
a) \(S=1+3+3^2+3^3+...3^{99}\)
\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)
\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)
\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)
\(\Rightarrow dpcm\)
b) \(S=1+4+4^2+4^3+...4^{62}\)
\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)
\(\Rightarrow S=21+4^3.21+...4^{60}.21\)
\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)
\(\Rightarrow dpcm\)
LM
25 tháng 10 2021
a, 6100 - 1 = (6 . 6 . 6 ..... 6) - 1 = [(...6) . (...6) . (...6) ..... (...6)] - 1 = (...6) - 1 = ...5 \(⋮\) 5
LM
25 tháng 10 2021
b, 2120 - 1110 = (21 . 21 . 21 . 21 . 21..... 21) - (11 . 11 . 11 . 11 ..... 11) = [(...1) . (...1) . (...1) . (...1).....(...1)] - [(...1) . (...1) . (...1) . (...1).....(...1)] = (...1) - (...1) = ....0 \(⋮\) 2; \(⋮\) 5
13 tháng 11 2023
1: \(A=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{97}\right)\)
\(=30\left(1+2^4+...+2^{96}\right)⋮30\)
2:
\(B=3+3^2+3^3+...+3^{2022}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2021}+3^{2022}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{2020}\left(3+3^2\right)\)
\(=12\left(1+3^2+...+3^{2020}\right)⋮12\)
A= 1+3+32+33+...+399
A= (1+3+32+33)+...+(396+397+398+399)
A= (1+3+32+33)+...+396(1+3+32+33)
A= 40 + ... + 399.40
Vì 40 chia hết cho 40 nên A chia hết cho 40
Chúc bn học tốt
\(A=1+3+3^2+...+3^{99}=\left(1+3+3^2+3^3\right)+...+3^{96}\left(1+3+3^2+3^3\right)\)
\(=40+...+3^{99}.40=40\left(1+3^{99}\right)⋮40\)
Vậy ta có đpcm