\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4.\left(3+3^3+...+3^{2009}\right)\)

⇒ \(B\) ⋮ 4

b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)

`#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 \(A\vdots13\)

\(\text{#}Toru\)

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:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

đúng rùi

\(S=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\\ =\left(3+3^2+3^3\right)+3^3.\left(3+3^2+3^3\right)+3^6.\left(3+3^2+3^3\right)\\ =39+3^3.39+3^6.39\\ =-39.\left(-1-3^3-3^6\right)⋮\left(-39\right)\)

S = 3 + 3² + 3³ + 3⁴ + 3⁵ + 3⁶ + 3⁷ + 3⁸ + 3⁹

S = ( 3 + 3² + 3³ ) +3⁴ + 3⁵ + 3⁶ + 3⁷ + 3⁸ + 3⁹ 

S = 39 + 3⁴ + 3⁵ + 3⁶ + 3⁷ + 3⁸ + 3⁹

Vì 39 ⋮ -39

<=> S ⋮ -39

a: \(B=3^1+3^2+...+3^{2010}\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4\left(3+3^3+...+3^{2009}\right)⋮4\)

\(B=3\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)

\(=13\left(3+...+3^{2008}\right)⋮13\)

b: \(C=5^1+5^2+...+5^{2010}\)

\(=5\left(1+5\right)+...+5^{2009}\left(1+5\right)\)

\(=6\left(5+...+5^{2009}\right)⋮6\)

\(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)\)

\(=31\left(5+...+5^{2008}\right)⋮31\)

c: \(D=7\left(1+7\right)+...+7^{2009}\left(1+7\right)\)

\(=8\left(7+...+7^{2009}\right)⋮8\)

\(D=7\left(1+7+7^2\right)+...+7^{2008}\left(1+7+7^2\right)\)

\(=57\left(7+...+7^{2008}\right)⋮57\)

\(A=3+3^2+3^3+...+3^{2012}\\ A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\\ A=120+...+3^{2008}.120\\ A=120.\left(1+...+3^{2008}\right)⋮120\)

\(=\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)

\(=120+...+3^{2008}.120=120\left(1+...+3^{2008}\right)⋮120\)

Đặt A = 3² + 3³ + 3⁴ + ... + 3⁹⁹

= 3² + 3³ + (3⁴ + 3⁵ + 3⁶) + (3⁷ + 3⁸ + 3⁹) + ... + (3⁹⁷ + 3⁹⁸ + 3⁹⁹)

= 36 + 3⁴.(1 + 3 + 3²) + 3⁷.(1 + 3 + 3²) + ... + 3⁹⁷.(1 + 3 + 3²)

= 36 + 3⁴.13 + 3⁷.13 + ... + 3⁹⁷.13

= 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷)

Do 36 không chia hết cho 13

13.(3⁴ + 3⁷ + ... + 3⁹⁷) ⋮ 13

⇒ 36 + 13.(3⁴ + 3⁷ + ... + 3⁹⁷) không chia hết cho 13

⇒ A không chia hết cho 13

Em xem lại đề nhé, có thể em viết thiếu số 3 rồi

\(B=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\\=(3+3^2)+(3^3+3^4)+(3^5+3^6)+(3^7+3^8)\\=3\cdot(1+3)+3^3\cdot(1+3)+3^5\cdot(1+3)+3^7\cdot(1+3)\\=3\cdot4+3^3\cdot4+3^5\cdot4+3^7\cdot4\\=4\cdot(3+3^3+3^5+3^7)\)

Vì \(4\cdot(3+3^3+3^5+3^7) \vdots 4\)

nên \(B\vdots4\).

`#3107.101107`

\(B=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+\left(3^7+3^8\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+3^5\left(1+3\right)+3^7\left(1+3\right)\)

\(=\left(1+3\right)\left(3+3^3+3^5+3^7\right)\)

\(=4\left(3+3^3+3^5+3^7\right)\)

Vì \(4\left(3^3+3^5+3^7\right)\) $\vdots 4$

`\Rightarrow B \vdots 4`

Vậy, `B \vdots 4.`

B=3+32+33+34+35+36+37+38=(3+32)+(33+34)+(35+36)+(37+38)=3⋅(1+3)+33⋅(1+3)+35⋅(1+3)+37⋅(1+3)=3⋅4+33⋅4+35⋅4+37⋅4=4⋅(3+33+35+37)

Vì 4⋅(3+33+35+37)⋮44⋅(3+33+35+37)⋮4

nên $B⋮4$.