Question

[Phan Hải Đăng]

Cho C= 1+3+3^2+...+3^11. Chứng minh:

a/ C chia hết cho 13

b/ C chia hết cho 40

Answer 1

[Han Sara ft Tùng Maru]

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

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) \(C=1+3+3^2+...+3^{11}\)

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

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

      \(=40+3^4.40+3^8.40\)

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

\(\Rightarrow C⋮40\)

Answer 2

[Lê Bảo Sơn]

C chia het cho ca 13 va 40