Question

chứng minh rằng : A=n³+3n²-n-3 chia hết cho 8 Vn thuộc N. n lẻ

Answer 1

\(A=n^3+3n^2-n-3=n^2\left(n+3\right)-\left(n+3\right)=\left(n^2-1\right)\left(n+3\right)\)

\(=\left(n-1\right)\left(n+1\right)\left(n+3\right)\)

Do n lẻ nên n=2k+1 (k thuộc N)

=>\(A=\left(n-1\right)\left(n+1\right)\left(n+3\right)=\left(2k+1-1\right)\left(2k+1+1\right)\left(2k+1+3\right)\)

\(=2k\left(2k+2\right)\left(2k+4\right)=2k.2\left(k+1\right).2\left(k+2\right)=8k\left(k+1\right)\left(k+2\right)\) chia hết cho 8

Vậy ta có đpcm