Ta có : \(3+3^2+3^3+3^4+....+3^{25}\)
\(=\left(3+3^3+3^5+....+3^{25}\right)+\left(3^2+3^4+3^6+....+3^{24}\right)\)
Đặt \(A=3+3^3+3^5+....+3^{25}\)
\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{21}+3^{23}+3^{25}\right)\)
\(=1.\left(3+3^3+3^5\right)+3^6.\left(3+3^3+3^5\right)+....+3^{20}.\left(3+3^3+3^5\right)\)
\(=1.273+3^6.273+....+3^{20}.273\)
\(=273.\left(1+3^6+...+3^{20}\right)\)\(\Rightarrow A⋮273\) (1)
Đặt \(B=3^2+3^4+3^6+...+3^{24}\)
\(=\left(3^2+3^4+3^6\right)+\left(3^8+3^{10}+3^{12}\right)+...+\left(3^{20}+3^{22}+3^{24}\right)\)
\(=3.\left(3+3^3+3^5\right)+3^7.\left(3+3^3+3^5\right)+...+3^{19}.\left(3+3^3+3^5\right)\)
\(=3.273+3^7.273+....+3^{19}.273\)
\(=273.\left(3+3^7+...+3^{19}\right)\) \(\Rightarrow B⋮273\) (2)
Từ (1) và (2), suy ra : \(A+B⋮273\)
Vậy \(3+3^2+3^3+....+3^{25}\) chia hết cho 273
~ Học tốt ~
