Question

Demonstrați ca 1+3^4+3^8+3^12+3^16 este divizibil cu 5.Generalizati.​

Answer 1

Ai mai jos solutia.

Answer 2

$\displaystyle\bf\\\boxed{\bf U(3^{4k})=\overline{...1},~\forall~k\in\mathbb{N} }~.\\\\demonstratie~:~ 3^{4k}=81^k,~deci~U(81^k)=1^k=1.\\\\\boxed{\bf un~numar~este~divizibil~cu~5,~daca~ultima~sa~cifra~este~0~sau~5}\\\\----------------------------\\\\1+3^4+3^8+3^{12}+3^{16},~observam~ca~exponentii~sunt~de~forma~4k \implies\\U(1+3^4+3^8+3^{12}+3^{18})=1+1+1+1+1=5.\\\\\boxed{\bf GENERALIZARE}~:~\\\\Demonstrati~ca~1+\underbrace{3^4+3^8+3^{12}...+3^{4k}}_\text{de~4m~ori~sau~9m~ori.}~este~multiplu~de~5.$

$\displaystyle\bf\\\boxed{\bf DEMONSTRATIE}~:~\\\\cazul~1~:~avem~4m~termeni~in~suma.\\U(1+4(U(3^k)))= 5~\vdots~5,~din~cauza~ultimei~cifre.\\cazul~2~:~avem~9m~termeni~in~suma.\\U(1+9(U(3^k)))=0~\vdots~5,~din~cauza~ultimei~cifre.$