\(3^{n+1}+3^{n+2}+...+3^{n+100}=\left(3^{n+1}+3^{n+2}+3^{n+3}+3^{n+4}\right)+...+\left(3^{n+97}+3^{n+98}+3^{n+99}+3^{n+100}\right)=3^n.\left(3+3^2+3^3+3^4\right)+...+3^{n+96}.\left(3+3^2+3^3+3^4\right)=3^n.120+...+3^{n+96}.120=\left(3^n+...+3^{n+96}\right).120\)
chia hết cho 120