calibudaicho

C/M C\(⋮\)4

\(C=1+3+3^2+...+3^{99}⋮4\)

\(C=\left(1+3\right)+\left(3^2+3^3\right)+...+\left(3^{98}+3^{99}\right)⋮4\)

\(C=\left(1+3\right)+3^2.\left(1+3\right)+...+3^{98}.\left(1+3\right)⋮4\)

\(C=4+3^2.4+...+3^{98}.4⋮4\)

\(C=4.\left(1+3^2+...+3^{98}\right)⋮4\)

C/M C\(⋮\)40

\(C=1+3+3^2+...+3^{99}⋮40\)

\(C=\left(1+3+3^2+3^3\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)⋮40\)

\(C=\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)⋮40\)

\(C=40.1+...+3^{96}.40⋮40\)

\(C=40.\left(1+...+3^{96}\right)⋮40\)

\(C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+......+\left(3^9+3^{10}+3^{11}\right)\)

\(C=13.1+3^3.13+......+3^9.13\)

\(C=13.\left(1+3^3+3^6+3^9\right)\)

Chia hết cho 13

\(C=\left(1+3+3^2+3^3\right)+......+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(C=40.1+40.3^4+40.3^8\)

\(C=40.\left(1+3^4+3^8\right)\)

Chia hết cho 40

Cho A = 1-3+3 mũ 2-3 mũ 3+3 mũ 4-3 mũ 5+.....+3 mũ 98-3 mũ 99 chứng to A chia hết cho 20

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C=1+3+3^2+...+3^11

C=(1+3+3^2)+...+(3^9+3^10+3^11)

C=13+13.3^3+...+13.3^9

C=13(1+3^3+3^6+3^9) chia hết 13

C=1+3+3^2+...+3^11

C=(1+3+3^2+3^3)+...+(3^8+3^9+3^10+3^11)

C=40+40.3^4+40.3^8

=40(1+3^4+3^8) chia hết 40

Ta có : 3C = 3 + 3^2 + 3^3 + ...3^12
=> 3C - C = (3 + 3^2 + 3^3 + ...3^12) - (1+3+3^2+3^3+....+3^11) = 3^12 - 1 = 531440
hay 2C = 531440 => C = 53144 :2 = 265720

265720 = 20440.13 => C chia hết cho 13 ( vì có thừa số 13)

265720 = 6643.40 => C chia hết cho 40 ( vì có thừa số 40)

\(C=1+3+3^2+...+3^{11}\)

a) \(C=1+3+3^2+...+3^{11}\)

       \(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\)

       \(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+3^9\left(1+3+3^2\right)\)

      \(=13+3^3.13+3^6.13+3^9.13\)

      \(=13\left(1+3^3+3^6+3^9\right)⋮13\)

\(\Rightarrow C⋮13\)

b) \(C=1+3+3^2+...+3^{11}\)

       \(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)

       \(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)

      \(=40+3^4.40+3^8.40\)

      \(=40\left(1+3^4+3^8\right)⋮40\)

\(\Rightarrow C⋮40\)

C chia het cho ca 13 va 40
 

a) \(C=1+3+3^2+........+3^{11}\)

\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...........+\left(3^9+3^{10}+3^{11}\right)\)

\(=\left(1+3+3^2\right)+3^3.\left(1+3+3^2\right)+.....+3^9.\left(1+3+3^2\right)\)

\(=\left(1+3+3^2\right).\left(1+3^3+.......+3^9\right)\)

\(=13.\left(1+3^3+.......+3^9\right)⋮13\)( đpcm )

b) \(C=1+3+3^2+.......+3^{11}\)

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(=\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+3^8.\left(1+3+3^2+3^3\right)\)

\(=\left(1+3+3^2+3^3\right).\left(1+3^4+3^8\right)\)

\(=40.\left(1+3^4+3^8\right)⋮40\)( đpcm )

C=(1+3+3²)+(3³+3⁴+3⁵)+...+(3⁹+3¹⁰+3¹¹)

C=13+3³(1+3+3²)+...+3⁹(1+3+3²)

C=13+3³.13+...+3⁹.13

C=13(1+3³+...+3⁹)

Vì nó có thừa số 13 nên chia hết cho 13 (1+3³+...+3⁹ là STN)

C=(1+3+3²+3³)+(3⁴+3⁵+3⁶+3⁷)+(3⁸+3⁹+3¹⁰+3¹¹)

C=40+3⁴(1+3+3²+3³)+3⁸(1+3+3²+3³)

C=40+3⁴.40+3⁸.40

=40(1+3⁴+3⁸)

=>C chia hết cho 40

a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4\right)+3^7\times\left(1+3^2+3^4\right)+...+3^{1987}\times\left(1+3^2+3^4\right)\)

\(=3\times91+3^7\times91+...+3^{1987}\times91\)

\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)

\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)

Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.

b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)

\(=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)

\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)

\(=3\times820+...+3^{1985}\times820\)

\(=3\times20\times41+...+3^{1985}\times20\times41\)

\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)

Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.