Ta có: 3^0 + 3^1 + 3^2 + 3^3 + ... + 3^11
= ( 3^0 + 3^1 + 3^2 + 3^3 ) + ... + ( 3^8 + 3^9 + 3^10 + 3^11 )
= 40 + ... + 3^8 . ( 3^0 + 3^1 + 3^2 + 3^3 )
= 40 + ... + 3^8 . 40
= 40 . ( 1 + ... + 3^8 ) \(⋮\)40
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\(1+3+3^2+............+3^{11}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(=1\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)
\(=1.40+3^4.40+3^8.40\)
\(=40\left(1+3^4+3^8\right)⋮40\left(đpcm\right)\)