Nên \(3^1+3^2+3^3+3^4+3^5+...+3^{2012}=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)=120+3^4.120+...+3^{2008}.120=120.\left(1+3^4+...+3^{2008}\right)⋮120\) Vậy tổng \(S=3^1+3^2+3^3+...+3^{2012}⋮120\)