Question

Domonstrați că numarul a=2^ n+2 • 3 ^n+2 + 5 • 6 ^n+ 2^ n+1 • 3^ n+1

Answer 1

$a=2^{n+2}\cdot3^{n+2}+5\cdot6^n+2^{n+1}\cdot3^{n+1} \\ \\ =6^{n+2}+5\cdot6^n+6^{n+1} \\ \\ =6^n\cdot36+6^n\cdot5+6^n\cdot6 \\ \\ =6^n(36+5+6) \\ \\ =6^n\cdot47$
Care este divizibil cu 47, daca asta trebuia sa demonstrez