Bài 1:\(A=3+3^2+...+3^{100}\)=>\(3\cdot A=3^2+3^3+...+3^{101}\)=>\(3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)=>\(2A=3^{101}-3\)=>\(2A+3=3^{101}\)mà \(2A+3=3^n\)nên n=101 Bài 2:a: \(M=3+3^2+3^3+3^4+...+3^{100}\)\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)\(=12\left(1+3^2+...+3^{98}\right)⋮12\)=>\(M=4\cdot3\cdot\left(1+3^2+...+3^{98}\right)⋮4\)b: \(M=3+3^2+...+3^{100}\)=>\(3M=3^2+3^3+...+3^{101}\)=>\(3M-M=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)=>\(2M=3^{101}-3\)=>\(2M+3=3^{101}\)=>n=101Câu trả lời: Bài 1:\(A=3+3^2+...+3^{100}\)=>\(3\cdot A=3^2+3^3+...+3^{101}\)=>\(3A-A=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)=>\(2A=3^{101}-3\)=>\(2A+3=3^{101}\)mà \(2A+3=3^n\)nên n=101 Bài 2:a: \(M=3+3^2+3^3+3^4+...+3^{100}\)\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{99}+3^{100}\right)\)\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{98}\left(3+3^2\right)\)\(=12\left(1+3^2+...+3^{98}\right)⋮12\)=>\(M=4\cdot3\cdot\left(1+3^2+...+3^{98}\right)⋮4\)b: \(M=3+3^2+...+3^{100}\)=>\(3M=3^2+3^3+...+3^{101}\)=>\(3M-M=3^2+3^3+...+3^{101}-3-3^2-...-3^{100}\)=>\(2M=3^{101}-3\)=>\(2M+3=3^{101}\)=>n=101