ta có :
\(A=3+3^2+3^3+..+3^{60}=\left(3+3^2\right)+\left(3^3+3^4\right)+..+\left(3^{59}+3^{60}\right)\)
\(=3.4+3^3.4+..+3^{59}.4\text{ nên A chia hết cho 4}\)
mà : \(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+..+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3.13+3^4.13+3^7.13+..+3^{58}.13\text{ nên A chia hết cho 13}\)
Đặt :
\(A=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=3.4+3^3.4+...+3^{59}.4\)
\(=4\left(3+3^3+...+3^{59}\right)\)
Vì \(4⋮4\)
\(\Rightarrow4\left(3+3^3+...+3^{59}\right)⋮4\)
\(\Rightarrow3+3^2+3^3+...+3^{60}⋮4\)
Đặt :
\(B=3+3^2+3^3+...+3^{60}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=3.13+...+3^{58}.13\)
\(=13\left(3+3^4+...+3^{58}\right)\)
Ta có : \(13⋮13\)
\(\Rightarrow13\left(3+3^4+...+3^{58}\right)⋮13\)
\(\Rightarrow3+3^2+3^3+...+3^{60}⋮13\)