\(\left(2n-1\right)^3-\left(2n-1\right)\)
\(=\left(2n-1\right)\left[\left(2n-1\right)^2-1^2\right]\)
\(=\left(2n-1\right)\left(2n-1-1\right)\left(2n-1+1\right)\)
\(=\left(2n-1\right)\left(2n-2\right)\left(2n\right)\)
Vì 2n và 2n - 2 là 2 số chắn liên tiếp nên có tích chia hết cho 8
=>\(\left(2n-1\right)\left(2n-2\right)\left(2n\right)\) chia hết cho 8
=>\(\left(2n-1\right)^3-\left(2n-1\right)\) chia hết cho 8 (đpcm)
\(\left(2n-1\right)^3-\left(2n-1\right)\)
\(=2n^3-1^3-2n-1\)
\(=\left(2n^3-2n\right)-\left(1^3-1\right)\)
\(=\left(2n^3-2n\right)-1^3+1\)
\(=\left(2n^3-2n\right)-2\)
\(=\left(2n.2n.2n-2n\right)-2\)
\(=\left(8n_{ }^3-2n\right)-2\)
\(=\left(-2.4+8\right)n\)
\(=\left(-8+8\right)n\)
\(=0n⋮8\)
Vậy ...
