Răspuns
Explicație pas cu pas:
[tex]P(n):1+2+2^2+\ldots+2^{n}=2^{n+1}-1,n\in\mathbb{N}\\\text{Etapa 1:Verificarea}\\P(0): 1=2^{0+1}-1\\~~~~~~~~~ 1=2-1\\~~~~~~~~~ 1=1,~adevarat\\\text{Etapa 2:Demonstratia propriu-zisa}\\\text{Presupunem P(k) adevarat pentru orice k}\in\mathbb{N}.\text{Se demonstreaza ca P(k+1) }\\ \text{este adevarat}\\P(k):1+2+2^2+\ldots+2^k=2^{k+1}-1,k\in\mathbb{N}\\P(k+1):1+2+2^2+\ldots+2^{k+1}=2^{k+2}-1\\~~~~~~~~~~~~~~~ 1+2+2^2+\ldots+2^{k}+2^{k+1}=2^{k+2}-1\\~~~~~~~~~~~~~~~ 2^{k+1}-1+2^{k+1}=2^{k+2}-1[/tex]
[tex]\displaystyle ~~~~~~~~~~~~~ 2\cdot 2^{k+1}=2^{k+2}\\~~~~~~~~~~~~~ 2^{k+2}=2^{k+2},~adevarat .\Rightarrow P(k)-adevarat\Rightarrow P(n)-adevarat\forall n\in\mathbb{N}[/tex]
