\(A=n^3+3n^2+5n+3\)
\(=n^2\left(n+1\right)+2n\left(n+1\right)+3\left(n+1\right)\)
\(=\left(n+1\right)\left(n^2+2n+3\right)\)
\(=\left(n+1\right)\left[n\left(n+2\right)+3\right]\)
\(=n\left(n+1\right)\left(n+2\right)+3\left(n+1\right)\)
Do n ; n + 1 ; n + 2 là 3 số nguyên dương liên tiếp
\(\Rightarrow n\left(n+1\right)\left(n+2\right)⋮3\)
\(\Rightarrow...+3\left(n+1\right)⋮3\)
hay \(A⋮3\left(đpcm\right)\)
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\(A=n^3+3n^2+6n-\left(n+3\right)+6\)
\(=\left(n^2-1\right)\left(n+3\right)+6n+6\)
\(=\left(n-1\right)\left(n+1\right)\left(n+3\right)+6\left(n+1\right)\)
Có: \(n+3\equiv n\)(mod 3)
mà \(\left(n-1\right)n\left(n+1\right)⋮3\forall n\in Z^+\)
nên \(A⋮3\forall n\in Z^+\)
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