Question

Vă rog frumos sa răspundeți corect și repede.

Answer 1

Vom demonstra prin inductie.
Vom nota orice multiplu de 17 cu M17. Astfel:
M17 + M17 = M17
M17 * M17 = M17
a * M17 = M17, pentru orice a ∈ Z

Fie:
$P(n):3\cdot5^{2n+1}+2^{3n+1}=\mathcal{M}_{17}\ \ , \ n\in N$

$P(0)=3\cdot5^1+2^1=17=\mathcal{M}_{17}\ ,\ \text{ceea ce e adevarat}$

[tex]\text{Presupunem ca } P(k) \text{ este o propozitie adevarata:}\\ P(k):3\cdot5^{2k+1}+2^{3k+1}=\mathcal{M}_{17}\ \ \text{(adevarat)}[/tex]

Acum trebuie sa demonstram ca P(k + 1) este adevarata, tinand cont ca P(k) este adevarata:

[tex]P(k+1):3\cdot5^{2k+3}+2^{3k+4}=\mathcal{M}_{17}\ \ \text{(de demonstrat)}\\\\ 3\cdot5^{2k+1}+2^{3k+1}=\mathcal{M}_{17}\rightarrow2^{3k+1}=\mathcal{M}_{17}-3\cdot5^{2k+1}\\\\ P(k+1):3\cdot5^2\cdot5^{2k+1}+2^3\cdot2^{3k+1}\\ P(k+1):75\cdot5^{2k+1}+8(\mathcal{M}_{17}-3\cdot5^{2k+1})\\ P(k+1):5^{2k+1}\cdot75-5^{2k+1}\cdot24+8\mathcal{M}_{17}\\ P(k+1):5^{2k+1}(75-24)+\mathcal{M}_{17}\\ P(k+1):2^{2k+1}\cdot51+\mathcal{M}_{17}=17(2^{2k+1}\cdot3)+\mathcal{M}_{17}=\mathcal{M}_{17}\ \ \text{q.e.d.}[/tex]

Asta inseamna ca P(n) este adevarata pentru orice n natural ≥ 0