Question

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

Answer 1

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

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

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

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

Answer 2

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

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

Answer 3

⇔M=(2+22+23+24)+(25+26+27+28)+...+(217+218+219+220)

 

⇔M=30+24(2+22+23+24)+...+216(2+22+23+24)

 

⇔M=30+24.30+...+216.30

 

⇔M=30(1+24+...+216)⋮5

Answer 4

28)+...+(217+218+219+220)

⇔M=30+24(2+22+23+24)+...+216(2+22+23+24)⇔M=30+24(2+22+23+24)+...+216(2+22+23+24)

⇔M=30+24.30+...+216.30⇔M=30+24.30+...+216.30

⇔M=30(1+24+...+216)5⇔M=30(1+24+...+