\(A=1+3+3^2+3^3+......+3^{99}\\ =\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+...+\left(3^{96}+3^{97}+3^{98}+3^{99}\right)\\ =40+3^4\left(1+3+3^2+3^3\right)+....+3^{96}\left(1+3+3^2+3^3\right)\\ =40+3^4.40+.....+3^{96}.40\\ =40\left(1+3^4+....+3^{96}\right)⋮40\)
Chứng tỏ rằng tổng \(1+3+3^2+.....+3^{99}\)chia hết cho 40
=> \(1+3+3^2+.....+3^{99}\)
= \(3^0+3^1+3^2+.......+3^{99}\)
= \(\left(3^0+3^1+3^2+3^3\right)+\left(3^4+3^5+.....+3^{99}\right)\)
=\(3^0.\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+...+3^{95}\right)\)
=\(3^0.40+3^4.40+...+3^{95}\)
= 40. \(\left(3^0+3^4\right)+.....+3^{95}\)
Vậy 40. \(\left(3^0+3^4\right)+.....+3^{95}\)\(⋮\) 40
