Question

Cho A =7 + 72 + 73 + ... + 7119 + 7120. Chứng minh chia hết cho 57

Answer 1

\(A=7+7^2+7^3+...+7^{120}\\ A=\left(7+7^2+7^3\right)+...+\left(7^{118}+7^{119}+7^{120}\right)\\ A=7\times\left(1+7+7^2\right)+...+7^{118}\times\left(1+7+7^2\right)\\ A=7\times57+7^4\times57+...+7^{118}\times57\\ A=57\times\left(7+7^4+...+7^{118}\right)\\ \Rightarrow A⋮57\)

Answer 2

A = 7 + 72 + 73 + ... + 7119 + 7120

A = (71 + 72 + 73) + (74 + 75 + 76) + ... + (7118 + 7119 + 7120)

A = 7(1 + 7 + 72) + 74(1 + 7 + 72) + ... + 7118(1 + 7 + 72)

A = 7.57 + 74.57 + ... + 7118.57

A = 57(7 + 74 + ... + 7118)

Vì 57 ⋮ 57 nên 57(7 + 74 + ... + 7118) ⋮ 57