Question

Demonstrati ca numarul A=63(la puterea N ) + 7(la puterea N+1) * 3(la puterea 2n+1) - 21(la puterea N) * 3(la puterea n+2)      n apartine numerelor naturale este divizibil cu 13 !

Answer 1

$63^n+7^{n+1}\cdot3^{2n+1}+21^n\cdot3^{n+2}=63^n+7^n\cdot7\cdot(3^2)^n\cdot3+21^n\cdot3^n\cdot3^2=$

$=63^n+63^n\cdot3+63^n\cdot9=63^n(1+3+9)=63^n\cdot13\ \vdots\ 13$