\(A=3+3^2+3^3+3^4+...+3^9+3^{10}\)(có 10 số)
\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^9+3^{10}\right)\)(có 5 nhóm)
\(A=3\left(1+3\right)+3^3\left(1+3\right)+...+3^9\left(1+3\right)\)
\(A=\left(1+3\right)\left(3+3^3+...+3^9\right)\)
\(A=4\left(3+3^3+...+3^9\right)⋮4\left(đpcm\right)\)
A = 3+32+33+...+39+310
A = (3+ 32)+(33+34)+...+(39+310)
A = 3(1+3)+33(1+3)+...+39 (1+3)
A = (1+3)(3+33+...+39)
A = 4(3+33+...+39) => chia hết cho 4