\(A=1+3+3^2+3^3+....+3^{11}\)
\(=\left(1+3+3^2\right)\left(3^3+3^4+3^5\right)+.....+\left(3^9+3^{10}+3^{11}\right)\)
\(=13.1+3^3.13+...+3^9.13\)
\(=13.\left(1+3^3+3^6+3^9\right)\)
Vì có cơ số là 13 => A chia hết cho 13
b) \(A=1+3+3^2+3^3+....+3^{11}\)
\(=40.1+40.3^4+40.3^8\)
\(=40.\left(1+3^4+3^8\right)\)
Vì có cơ số 40 nên A chia hết 40
Ta có
\(\left(+\right)A=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+.....+3^9\left(1+3+3^2\right)=13\left(1+3^3+...+3^9\right)\)(chia hết cho 13)
\(\left(+\right)A=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)=40\left(1+3^4+3^8\right)\) chia hết cho 40
)