Question

Đố ai lm đc câu này đấy:

Chứng minh rằng: \(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}⋮120\left(x\in Z\right)\)

 

Answer 1

\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)

\(=3^x\left(3^1+3^2+3^3+3^4\right)+...+3^{x+96}\left(3^1+3^2+3^3+3^4\right)=3^x.120+3^{x+4}.120+...+3^{x+96}.120=120\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)