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cho A = 1 + 3 + 32 + 33 + ... + 311
a ) chứng minh A chia hết cho 13
b) chứng minh A chia hết cho 40
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
Câu 1:
$A=(2+2^2)+(2^3+2^4)+(2^5+2^6)+....+(2^{2019}+2^{2020})$
$=2(1+2)+2^3(1+2)+2^5(1+2)+....+2^{2019}(1+2)$
$=(1+2)(2+2^3+2^5+...+2^{2019})=3(2+2^3+2^5+...+2^{2019})\vdots 3$
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$A=2+(2^2+2^3+2^4)+(2^5+2^6+2^7)+....+(2^{2018}+2^{2019}+2^{2020})$
$=2+2^2(1+2+2^2)+2^5(1+2+2^2)+....+2^{2018}(1+2+2^2)$
$=2+(1+2+2^2)(2^2+2^5+....+2^{2018})$
$=2+7(2^2+2^5+...+2^{2018})$
$\Rightarrow A$ chia $7$ dư $2$.
Câu 2:
$B=(3+3^2)+(3^3+3^4)+....+(3^{2021}+3^{2022})$
$=3(1+3)+3^3(1+3)+...+3^{2021}(1+3)$
$=(1+3)(3+3^3+...+3^{2021})=4(3+3^3+....+3^{2021})\vdots 4$
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$B=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^{2020}+3^{2021}+3^{2022})$
$=3(1+3+3^2)+3^4(1+3+3^2)+....+3^{2020}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+...+3^{2020})=13(3+3^4+...+3^{2020})\vdots 13$ (đpcm)
\(A=3+3^2+3^3+3^4+.......+3^{100}\)
\(\Rightarrow A=\left(3+3^2+3^3+3^4\right)+.......+\left(3^{97}+3^{98}+3^{99}+3^{100}\right)\)
\(\Rightarrow A=3.\left(1+3+3^2+3^3\right)+........+3^{97}.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=3.40+.........+3^{97}.40\)
\(\Rightarrow A=40.\left(3+.......+3^{97}\right)\)
\(\Rightarrow A⋮40\)( 1 )
Vì \(A\)là tổng của các bậc lũy thừa của 3 nên \(A⋮3\)( 2 )
Từ ( 1 ) và ( 2 ) suy ra : \(A⋮40.3\)
\(\Rightarrow A⋮120\)
Vậy \(A⋮120\)( ĐPCM )
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 = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
\(B=3+3^2+3^3+...+3^{120}\)
Dễ thấy \(B\)chia hết cho \(3\)do là tổng của các số hạng chia hết cho \(3\).
\(B=3+3^2+3^3+...+3^{120}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{119}\right)⋮4\)
\(B=3+3^2+3^3+...+3^{120}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\)
\(=13\left(3+...+3^{118}\right)⋮13\)
a) \(B\)là tổng các số hạng chia hết cho \(3\)nên chia hết cho \(3\).
b) \(B=3+3^2+...+3^{120}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{119}\right)⋮4\)
c) \(B=3+3^2+...+3^{120}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{118}\left(1+3+3^2\right)\)
\(=13\left(3+3^4+...+3^{118}\right)⋮13\)
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
a) A luôn chia hết cho 3
A = (3 + 32) + (33 + 34) + ...+ (31997 + 31998) = 3.(1 + 3) + 33.(1 + 3) + ...+ 31997.(1 + 3) = 4.(3 + 33 + ...+ 31997)
=> A chia hết cho 4 ; A chia hết cho 3 => A chia hết cho 12
A = (3 + 32 + 33) + ...+ (31996 + 31997 + 31998) = 3.(1 + 3 + 32) + ...+ 31996.(1 + 3+ 32) = 13.(3 + 34 + ...+ 31996)
=> A chia hết cho 13. A chia hết cho 3 => A chia hết cho 39
b) A = (3 + 32 + 33 + 34) + ..+ (3997 + 3998 + 3999 + 31000)
A = 3.(1 + 3 + 32 + 33) + ...+ 3997.(1 + 3 + 32 + 33) = 40.(3 + ...+ 3997)
=> A chia hết cho 40 ; A chia hết cho 3
=> A chia hết cho 40.3 = 120
Vậy...
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