Toán Lớp 8: Chứng minh rằng: n³ + (n+1)³ + (n+2)³ chia hết cho 9 với mọi n ∈ N*

Question

Toán Lớp 8:  Chứng minh rằng: n³ + (n+1)³ + (n+2)³ chia hết cho 9 với mọi n ∈ N*

hothaibinh · Answer

Ta sẽ chứng minh n với $3$ dạng l&agrave; $3k;3k+1;3k+2$   *Với $n=3k$ :   =>n&sup3;+(n+1)&sup3;+(n+2)&sup3;=(3k)^3 + ( 3k +1 )^3  + (3k+2)&sup3;   V&igrave; (3k)^3  vdots 3 ; (3k+1)^3 chia $9$ dư $1$ ; $(3k+2)^3 $ chia $9$ dư $8$   =>(3k)^3 + ( 3k +1 )^3  + (3k+2)&sup3; vdots 9   =>n&sup3;+(n+1)&sup3;+(n+2)&sup3; vdots 9   *Với $n=3k+1$ :   =>n&sup3;+(n+1)&sup3;+(n+2)&sup3;=(3k+1)+(3k+2)^3 + (3k+3)^3    V&igrave; (3k+1)^3 chia $9$ dư $1$ ; $(3k+2)^3$ chia $9$ dư $8$ ; $(3k+3)^3  vdots 9$   =>(3k+1)+(3k+2)^3 + (3k+3)^3  vdots 9   =>n&sup3;+(n+1)&sup3;+(n+2)&sup3; vdots 9   *Với $n=3k+2$   =>n&sup3;+(n+1)&sup3;+(n+2)&sup3; vdots 9=(3k+2)^3 +(3k+3)^3 + (3k+4)^3   V&igrave; (3k+2)^3 chia $9$ dư $8$  ; $(3k+3)^3  vdots 9$ ; $(3k+4)^3 $ chia $9$ dư $1$   =>(3k+2)^3 +(3k+3)^3 + (3k+4)^3 vdots 9   =>n&sup3;+(n+1)&sup3;+(n+2)&sup3; vdots 9   Từ $3$ TH tr&ecirc;n =>n&sup3;+(n+1)&sup3;+(n+2)&sup3; vdots 9AAn&isin;NN*$(đpcm)$   Vậy n&sup3;+(n+1)&sup3;+(n+2)&sup3; vdots 9AAn&isin;NN