cho tớ hỏi CMR n(n+1)(2n+1) chia hết cho 2và 3
* Chứng minh chia hết cho 2:
-Nếu n chẵn: Đặt [tex]n=2k; k \in \mathbb{N}[/tex]
[tex]n(n+1)(2n+1)=2k(2k+1)(4k+1)\vdots 2[/tex]
-Nếu n lẻ: Đặt [tex]n=2k+1; k \in \mathbb{N}[/tex]
[tex]n(n+1)(2n+1)=(2k+1)(2k+1+1)(4k+2+1)=(2k+1)(2k+2)(4k+3)=(2k+1).2(k+1)(4k+3)\vdots 2[/tex]
* Chứng minh chia hết cho 3:
-Nếu n chia hết cho 3: Đặt [tex]n=3k; k \in \mathbb{N}[/tex]
[tex]n(n+1)(2n+1)=3k(3k+1)(6k+1)\vdots 3[/tex]
-Nếu n chia cho 3 dư 1: Đặt [tex]n=3k+1; k \in \mathbb{N}[/tex]
[tex]n(n+1)(2n+1)=(3k+1)(3k+1+1)(6k+2+1)=(3k+1)(3k+2)(6k+3)=(3k+1)(3k+2).3(2k+1)\vdots 3[/tex]
-Nếu n chia cho 3 dư 2: Đặt [tex]n=3k+2; k \in \mathbb{N}[/tex]
[tex]n(n+1)(2n+1)=(3k+2)(3k+2+1)(6k+4+1)=(3k+2)(3k+3)(6k+5)=(3k+2).3(k+1)(6k+5)\vdots 3[/tex]
[tex]\Rightarrow n(n+1)(2n+1)\vdots 2\; and \; \vdots 3[/tex]