Question

a, chứng minh: A=2^1+ 2^2+ 2^3+2^4+...+2^2010, chia hết cho 3 và 7

Answer 1

Ta có: \(A=2^1+2^2+2^3+...+2^{2010}\\ A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\\ A=3.\left(2+2^3+...+2^{2009}\right)⋮3\\ =>A⋮3->\left(a\right)\\ Ta-lại-có:A=2^1+2^2+2^3+...+2^{2010}\\ A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=2\left(1+2+4\right)+2^4\left(1+2+4\right)+2^{2008}\left(1+2+4\right)\\ A=7\left(2+2^4+2^7+2^{2008}\right)⋮7\\ =>A⋮7->\left(b\right)\\ Từ\left(a\right),\left(b\right)=>A⋮3;A⋮7\)