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`#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 = 32 + 33 + 34 +...+ 3101
A = 32.(1 + 3 + 32 + 33 +...+ 399)
A =32[(1+ 3+32+33) + (34+ 35+36+37)+...+ (396 + 397+ 398 + 399)
A = 32.[ 40 + 34.(1+ 3 + 32 + 33)+...+ 396.(1 + 3 + 32 + 33)
A = 32.[ 40 + 34. 40 + ...+ 396.40]
A = 32.40.[ 1 + 34+...+396]
A = 3.120.[1 + 34 +...+ 396]
120 ⋮ 120 ⇒ A = 3.120.[ 1 + 34 +...+396] ⋮ 120 (đpcm)
Các số hạng trong tổng \(A\) đều chia hết cho \(3\) nên \(\Rightarrow A⋮3\)
Vậy \(A⋮3\)
A=3+3^2+3^3+3^4+...+3^12
A=(3+3^2+3^3)+(3^4+3^5+3^6)+.....+(3^10+3^11+3^12) (gộp nhóm)
A=3.(1+3+3^2)+3^4.(1+3+3^2)+......+3^10.(1+3+3^2) (phân phối)
A=3.13+3^4.13+....+3^10.13
A=13.(3+3^4+....+3^10)
Vì 13⋮13
nên 13.(3+3^4+...+3^10)⋮13
=>A⋮13
\(A=1+3+3^2+...+3^{101}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{99}+3^{100}+3^{101}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{99}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+...+3^{99}\right)⋮13\)
a: \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3+3^2+...+3^{119}\right)⋮3\)
b: \(B=3+3^2+3^3+3^4+...+3^{2020}\)
\(=3\left(1+3\right)+...+3^{2019}\left(1+3\right)\)
\(=4\cdot\left(3+...+3^{2019}\right)⋮4\)
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
Ta có: S=32+33+34+...+3101
S=3(3+32+33+34)+35(3+32+33+34)+...+397(3+32+33+34)
S=(3+35+...+39)+(3+32+33+34)= M+120 => Schia hết cho 120
gọi tổng đó là S
S=3^2+3^3+3^4+...+3^101
S=3(3+3^2+3^3+...+3^100)
S=3[(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+...+(3^97+3^98+3^99+1^100)]
S=3[120 + 3^4(3+3^2+3^3+3^4)+....+3^96(3+3^2+3^3+3^4)]
S=3[120+3^4.120+...+3^96.120]
S=3.120(1+3^4+...+3^96] thì S chia hết 120