Question

Sa se arate ca: $2^{n}$·$15^{n+1}$+$6^{n}$·$5^{n+2}$-$10^{n}$·$3^{n+2}$≡31, ∀n∈N

Answer 1

$2^n\cdot 15^{n+1}+6^n\cdot 5^{n+2}-10^n\cdot 3^{n+2} = \\ \\=2^n\cdot 15^n\cdot 15+6^n\cdot 5^n\cdot 5^2-10^n\cdot 3^n\cdot 3^2 =\\ \\ = (2\cdot 15)^n\cdot 15 + (6\cdot 5)^n\cdot 5^2 - (10\cdot 3)^n\cdot 3^2 =\\ \\ = 30^{n}\cdot 15+30^n\cdot 5^2-30^n\cdot 3^2 = \\ \\ =30^n\cdot (15+5^2-3^2) = \\ \\ = 30^n\cdot 31\,\,\,\vdots\,\,\,31,\quad \forall n\in \mathbb{N}$

Answer 2

 

$\displaystyle\bf\\ \text{\bf~Sa se arate ca:}\\2^n\cdot15^{n+1}+6^n\cdot5^{n+2}-10^n\cdot3^{n+2}~\vdots~31\\\\\text{\bf Rezolvare:}\\\\2^n\cdot15^{n+1}+6^n\cdot5^{n+2}-10^n\cdot3^{n+2}=\\\\=2^n\cdot15^n\cdot15^1+6^n\cdot5^n\cdot5^2-10^n\cdot3^n\cdot3^2=\\\\=\Big(2\cdot15\Big)^n\cdot15+\Big(6\cdot5\Big)^n\cdot25-\Big(10\cdot3\Big)^n\cdot9=\\\\=30^n\cdot15+30^n\cdot25-30^n\cdot9=\\\\=30^n(15+25-9)=\\\\=30^n(40-9)=\\\\=30^n\cdot31~\vdots~31~~\text{\bf~deoarece un factor = 31.}$