Chứng minh:
[tex]1^{1991}+2^{1991}+3^{1991}+...+1991^{1991}[/tex] chia hết cho 11
@hdiemht 2 giúp em...
Ta có: [tex]a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}.b^2−...+a^2b^{n-3}-a.b^{n-2}+b^{n-1}) \forall n\in \mathbb{N^*}[/tex] và $n$ lẽ
Nên: [tex](a^n+b^n)\vdots (a+b)[/tex] [tex]\forall n\in \mathbb{N^*}[/tex] , $n$ lẽ
[tex]1^{1991}+2^{1991}+3^{1991}+...+1991^{1991}=1^{1991}+2^{1991}+3^{1991}+...+1990^{1991}+1991^{1991}=\left [ (1^{1991}+1990^{1991})+(2^{1991}+1989^{1991})+...+(995^{1991}+996^{1991})+1991^{1991} \right ]\vdots 1991[/tex]
Hay: $....$ chia hết cho $11$