Question

Sa se arate ca nr K=3^2n+3^2n+1+3^2n+2+3^2n+3+3^2n+4 este divizibil cu 11 pentru orice n € N

Answer 1

$k = 3^{2n}+3^{2n+1}+3^{2n+2}+3^{2n+3}+3^{2n+4} \\ \\ k = 3^{2n}+3^{2n}\cdot 3^1+3^{2n}+3^2+3^{2n}+3^3+3^{2n}+3^4 \\ \\ k = 3^{2n}\cdot \Big(1+3^1+3^2+3^3+3^4\Big) \\ \\ k = 3^{2n}\cdot \Big(1+3+9+27+81\Big) \\ \\ k = 3^{2n}\cdot (40+81) \\ \\ k = 3^{2n}\cdot 121\\ \\ k = 3^{2n}\cdot 11\cdot 11 \\ \\ \Rightarrow k \vdots 11,\quad \forall n\in \mathbb_{N}$

Answer 2

$k= 3^{2n}+ 3^{2n+1}+3^{2n+2}+ 3^{2n+4} =3^{2n}+3^{2n} *3^{1}+3^{2n} *3^{2}+3^{2n} *3^{4} \\ \\ k=3^{2n} (1+ 3^{1}+3^{2}+3^{3}+3^{4}) =3^{2n}*121=(3^{n}) ^{2}*11^{2}$

$(3^{n}) ^{2}*11*11$ este vizibil divizibil cu 11