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Kvothe
a fost răspuns

Demonstrati ca numarul 13^n+7^n-2 se divide cu 6, oricare ar fi n un numar natural.

Răspuns :



[tex] \boxed{(a+b)^{n}=M_{a}+b^{n}}[/tex]

[tex]13^{n}+7^{n}-2=\\
\\ =(6+7)^{n}+7^{n}-2=\\
\\ =M_{6}+7^{n}+7^{n}-2=\\
\\ =M_{6}+2\cdot7^{n}-2=\\
\\ =M_{6}+2(7^{n}-1)=\\
\\ =M_{6}+2[(6+1)^{n}-1]=\\
\\ =M_{6}+2( M_{6}+1^{n}-1)=\\
\\ =M_{6}+2\cdotM_{6}=\\
\\ =M_{6}(1+2)=\\
\\ =M_{6}\cdot3,~divizibil~cu~6[/tex]