Question

chứng minh 2 mũ 1 2 mũ 2 2 mũ 3 2 mũ 4 ....... 2 mũ 2016 chia hết cho 3 và 7

Answer 1

\(2^1+2^2+2^3+...+2^{2016}\)

\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2015}+2^{2016}\right)\)

\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2015}\left(1+2\right)\)

\(=3\left(2+2^3+...+2^{2015}\right)⋮3\)

\(2^1+2^2+2^3+...+2^{2016}\)

\(=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)\)

\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2014}\left(1+2+2^2\right)\)

\(=7\left(2+2^4+...+2^{2014}\right)⋮7\)