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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
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
\(C=1+3+3^2+3^3+...+3^{11}\\ a,C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\\ =13+3^3.\left(1+3+3^2\right)+3^6.\left(1+3+3^2\right)+3^9.\left(1+3+3^2\right)\\ =13+3^3.13+3^6.13+3^9.13\\ =13.\left(1+3^3+3^6+3^9\right)⋮13\)
Ý a phải chia hết cho 13 chứ em?
b: C=(1+3+3^2+3^3)+...+3^8(1+3+3^2+3^3)
=40(1+...+3^8) chia hết cho 40
a: C ko chia hết cho 15 nha bạn
\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
`#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.`
\(A=1+3+3^2+3^3+3^4+3^5+...+3^{101}\\=(1+3+3^2)+(3^3+3^4+3^5)+(3^6+3^7+3^8)+...+(3^{99}+3^{100}+3^{101})\\=13+3^3\cdot(1+3+3^2)+3^6\cdot(1+3+3^2)+...+3^{99}\cdot(1+3+3^2)\\=13+3^3\cdot13+3^6\cdot13+...+3^{99}\cdot13\\=13\cdot(1+3^3+3^6+...+3^{99})\)
Vì \(13\cdot(1+3^3+3^6...+3^{99}\vdots13\)
nên \(A\vdots13\)
\(\text{#}Toru\)
a: \(A=\left(1+3\right)+...+3^{10}\left(1+3\right)\)
\(=4\left(1+...+3^{10}\right)⋮4\)
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)
C=1+3+32+33+34+35+...+311
C=(1+3+32)+(33+34+35)+...+(39+310+311)
C=13+(33.1+33.3+33.32)+...+(39.1+39.3+39.32)
C=13+33.(1+3+32)+...+39.(1+3+32)
C=13.1+33.13+...+39.13
C=13.(1+33+35+37+39)\(⋮\)3
\(\Rightarrow\)C\(⋮\)3
Câu b ghép 4 số lại với nhau rồi làm như trên