Question

[Đỗ Đình Khoa]

CMR (1+7+7^2+7^3+...+7^101) chia hết cho 8

Answer 1

[Toru]

\(1+7+7^2+7^3+...+7^{101}\\=(1+7)+(7^2+7^3)+(7^4+7^5)+...+(7^{100}+7^{101})\\=8+7^2\cdot(1+7)+7^4\cdot(1+7)+...+7^{100}\cdot(1+7)\\=8+7^2\cdot8+7^4\cdot8+...+7^{100}\cdot8\\=8\cdot(1+7^2+7^4+...+7^{100})\)

Vì \(8\cdot\left(1+7^2+7^4+...+7^{100}\right)⋮8\)

\(\Rightarrowđpcm\)

Answer 2

[Nguyễn Lê Phước Thịnh]

\(1+7+7^2+7^3+...+7^{101}\)

\(=\left(1+7\right)+7^2\left(1+7\right)+...+7^{100}\left(1+7\right)\)

\(=8\left(1+7^2+...+7^{100}\right)⋮8\)