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

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 )

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

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

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

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

gffjjfhhhfh

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

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

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

\(=40.\left(1+3^4+3^8\right)⋮40\)

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

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)

Ta có : \(3C=3+3^2+3^3+......+3^{12}\)
\(\Rightarrow3C-C=\left(3+3^2+3^3+....+3^{12}\right)-\left(1+3+3^2+3^3+...+3^{11}\right)=3^{12}-1=531440\)
 \(hoặc\)\(2C=531140\Rightarrow C=265720\)chia hết cho 13 và 40

b, \(C=1+3+3^2+3^3+...+3^{11}\)

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

       \(=\left(1+3+9+27\right)+...+3^8.\left(1+3+3^2+3^3\right)\)

        \(=40+...+3^8.40\)

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

\(\Rightarrow\) \(C⋮40\)