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

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

\(=13\cdot\left(1+...+3^9\right)⋮13\)

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

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: Ta có: \(A=1+3+3^2+3^3+...+3^{2015}\)

\(=\left(1+3\right)+3^2\left(1+3\right)+...+3^{2014}\cdot\left(1+3\right)\)

\(=4\cdot\left(1+3^2+...+3^{2014}\right)⋮4\)

b: Ta có: \(A=1+3+3^2+3^3+...+3^{2015}\)

\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+...+3^{2013}\left(1+3+3^2\right)\)

\(=13\cdot\left(1+3^3+...+3^{2013}\right)⋮13\)

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

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

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

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

NHóm để đặt nhân tử có 13 và 40 nhen :3

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

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

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

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

\(\Rightarrow C⋮13\)

C =( 1 + 3 + 3^2) +( 3^3 + 3^4 + 3^5) + ...... + (3^9 + 3^10 + 3^11 )

C = 13.1 + 3^3 .13 + ...... + 3^9 .13

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

Chia hết cho 13

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

C = 40.1 + 40.3^4 + 40.3^8

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

Chia hết cho 40

Vậy......

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

    C = ( 1 + 3 + 3^2 ) + ( 3^3 + 3^4 + 3^5 ) + .... + ( 3^9 + 3^10 + 3^11 )

    C = 13  +  3^3 x ( 1 + 3 + 3^2 ) + ..... + 3^9 x ( 1 + 3 + 3^2 )

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

    C = 13 x ( 1 + 3^3 + ..... + 3^9 ) chia het cho 13.

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

    C = ( 1 + 3 + 3^2 + 3^3 ) + ( 3^4 + 3^5 + 3^6 + 3^7 ) + ( 3^8 + 3^9 + 3^10 + 3^11 )

    C = 40 + 3^4 x ( 1 +3 +3^2 + 3^3 ) + 3^8 x ( 1+ 3 + 3^2 + 3^3 )

    C = 40 x 3^4 x 40 + 3^8 x 40

    C = 40 x ( 1 + 3^4 + 3^8 ) chia het cho 40.

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

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

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

\(\Rightarrow\)C chia hết cho 13.

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

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

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

\(\Rightarrow\) C chia hết cho 40.

=> ĐPCM

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

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

\(=13+3^3.13+...+3^9.13\)

\(=13.\left(1+3^3+...+3^9\right)⋮13\)

\(\Rightarrow C⋮13\left(\text{đpcm}\right)\)

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

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

\(=40+3^4.40+3^8.40\)

\(=40.\left(1+3^4+3^8\right)⋮40\)

\(\Rightarrow C⋮40\left(\text{đpcm}\right)\)