Question

chứng minh: B= 3^1+3^2+3^3+3^4+...+3^2010 chia hết cho 4, 13

Answer 1

Ta có:

\(B=3^1+3^2+3^3+...+3^{2010}\\ B=\left(3^1+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\\ B=4\left(3+3^3+...+3^{2009}\right)⋮4\\ =>B⋮4->\left(a\right)\\ Ta-lại-có:B=3^1+3^2+3^3+...+3^{2010}\\ B=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=3\left(1+3+9\right)+3^4\left(1+3+9\right)+...+3^{2008}\left(1+3+9\right)\\ B=13\left(3+3^4+...+3^{2008}\right)⋮13\\ =>B⋮13->\left(b\right)\\ Từ\left(a\right),\left(b\right)=>B⋮4;B⋮13\)