Đáp án:
Giải thích các bước giải:
Nếu `n=2k`
`=>n(n+1)(2n+1)`
`=2k(2k+1)(4k+1)\vdots 2`
Nếu `n=2k+1`
`=>n(n+1)(2n+1)=(2k+1)(2k+2)(4k+3)=2(2k+1)(k+1)(4k+3)\vdots 2`
`=>` Vậy `n(n+1)(2n+1)` luôn chia hết cho 2
Nếu `n=3k`
`=>n(n+1)(2n+1)=3k(3k+1)(6k+1)\vdots 3`
Nếu `n=3k+1`
`=>n(n+1)(2n+1)=(3k+1)(3k+1+1)(6k+2+1)=3(3k+1)(3k+2)(2k+1)\vdots 3`
Nếu `n=3k+2`
`=>n(n+1)(2n+1)=(3k+2)(3k+2+1)(6k+4+1)=3(3k+2)(k+1)(6k+5)\vdots 3`
`=>n(n+1)(2n+1)` luôn chia hết cho 3
Ta có `n(n+1)(2n+1)` luôn chia hết cho 3 và 2
`=>n(n+1)(2n+1)` luôn chia hết `6`
`=>đpcm`
