A = 3 + 32 + 33 + 34 + ... + 312
A = (3 + 32 + 33 + 34) + (35 + 36 + 37 + 38) + (39 + 310 + 311 + 312)
A = 3.(1 + 3 + 32 + 33) + 35.(1 + 3 + 32 + 33) + 39.(1 + 3 + 32 + 33)
A = 3.120 + 35.120 + 39.120
A = 120.(3 + 35 + 39)
\(A=40.3.\left(3+3^5+3^9\right)⋮40\left(đpcm\right)\)
Ta có:
\(A=3+3^3+3^4+...+3^{12}\)
\(A=3.\left(1+2\right)+...+3^{11}\left(1+2\right)\)
\(A=3.3+...+3^{11}.3\)
\(\Rightarrow A⋮3\)
Vậy \(A⋮3\)
Không biết đúng ko nữa,lâu rồi ko hk 
A=3+3^2+3^3+3^4+...+3^12
A=(3+3^2)+(3^3+3^4)+(3^5+3^6)+(3^7+3^8)+(3^9+3^10)+(3^11+3^12)
A=3(1+3)+3^3(1+3)+3^5(1+3)+3^7(1+3)+3^9(1+3)+3^11(1+3)
A=3.4+3^3.4+3^5.4+3^7.4+3^9.4+3^11.4
A=4(3+3^3+3^5+3^7+3^9+3^11) chia hết cho 40
Vậy A chia hết cho 40.
A = 3 + 32 + 33 + 34 + ... + 312
A = ( 3 + 32 + 33 + 34 ) + ( 35 + 36 + 37 + 38 ) + ( 39 + 310 + 311 + 312 )
A = 3 ( 1 + 3 + 9 + 27 ) + 35 ( 1 + 3 + 9 + 27 ) + 39 ( 1 + 3 + 9 + 27 )
A = 3 . 40 + 35 . 40 + 39 . 40
A = 40 ( 3 + 35 + 39 ) \(⋮\) 40
=> A \(⋮\) 40
Ta có :
A = 3 + 32 + 33 + 34 + .... + 312
=> 3A =
