a) n2(n+1)+2n(n+1)
=(n+1)(n2+2n)
=n(n+1)(n+2)
ta thấy: n(n+1)⋮2 ∀ n (1)
n(n+1)(n+2)⋮3 ∀ n (2)
từ (1);(2)⇒n(n+1)(n+2)⋮6
hay n2(n+1)+2n(n+1)⋮6
b)(2n - 1)3 - 2n + 1
= (2n - 1)3 - (2n - 1)
= (2n - 1) [(2n - 1)2 - 12 ]
= (2n - 1) (2n - 1 + 1) (2n - 1 - 1)
= 2n (2n - 1) (2n - 2)
= 4n (2n - 1) (n - 1) ⋮ 4 (1)
mà (2n - 1) (n - 1) ⋮ 2 (2)
từ (1); (2) ⇒ (2n - 1)3 - 2n + 1 ⋮ 8 (x ∈ Z)
a) Ta có: \(n^2\left(n+1\right)+2n\left(n+1\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
b) Ta có: \(\left(2n-1\right)^3-2n+1\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1\right]\)
\(=\left(2n-1\right)\left(2n-2\right)\left(2n-3\right)⋮8\)