Question

[Khoa Tiến Đăng Nguyễn]

Cho M = 2 + 2^2 + 2^3 + … + 2^20. Chứng tỏ rằng M chia hết 5

Answer 1

[ILoveMath]

\(\Leftrightarrow M=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)

\(\Leftrightarrow M=30+2^4\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\)

\(\Leftrightarrow M=30+2^4.30+...+2^{16}.30\)

\(\Leftrightarrow M=30\left(1+2^4+...+2^{16}\right)⋮5\)

Answer 2

[Nguyễn Lê Phước Thịnh]

\(M=\left(2+2^2+2^3+2^4\right)+...+2^{17}\left(2+2^2+2^3+2^4\right)\)

\(=30\cdot\left(1+...+2^{17}\right)⋮5\)