Question

Chứng minh rằng với mọi n tích (n+1) . (n+4) chia hết cho 2

Answer 1

Vì \(n\in N\) nên ta có hai trường hợp.

TH1: n chẵn

\(\Rightarrow n=2k\left(k\in N\right)\)

\(\Leftrightarrow\left(n+1\right)\left(n+4\right)=\left(2k+1\right)\left(2k+4\right)\)\(=\left(2k+1\right)2\left(k+2\right)⋮2\)

\(\Rightarrow\left(n+1\right)\left(n+4\right)⋮2\)

TH2: n lẻ

\(\Rightarrow n=2k+1\left(k\in N\right)\)

\(\Leftrightarrow\left(n+1\right)\left(n+4\right)=\left(2k+2\right)\left(2k+5\right)=2\left(k+1\right)\left(2k+5\right)⋮2\)

\(\Rightarrow\left(n+1\right)\left(n+4\right)⋮2\)

Vậy \(\left(n+1\right)\left(n+4\right)⋮2\) với \(\forall n\in N\)