Question

Cho A=\(\text{1}^{\text{3}}+\text{2}^{\text{3}}+\text{3}^{\text{3}}+...+\text{16}^{\text{3}} \). Chứng minh rằng A⋮17

Answer 1

\(A=1^3+2^3+3^3+...+16^3\)

\(=\left(1^3+16^3\right)+\left(2^3+15^3\right)+\left(3^3+14^3\right)+...+\left(8^3+9^3\right)\)

\(=\left(1+16\right)\left(1^2-1.16+16^2\right)+\left(2+15\right)\left(2^2-2.15+15^2\right)+...+\left(8+9\right)\left(8^2-8.9+9^2\right)\)

\(=17.\left(1^2-1.16+16^2\right)+17.\left(2^2-2.15+15^2\right)+...+17.\left(8^2-8.9+9^2\right)\)

\(=17.\left(1^2-1.16+16^2+2^2-2.15+15^2+...+8^2-8.9+9^2\right)⋮17\)

hay : \(A⋮17\) ( đpcm )