Se rezolva prin inductie completa
n=0
2^0≥0+1 1≥1 evident
Presupui Pn adevarat. Se verifica daca Pn+1 adevarata
Pn=2^n≥n+1 relatia 1
Pn+1:2^(n+1)≥n+2
2^n·2≥n+2<=>
2^n+2^n≥n+1+1
2^n≥n+1 conf relatia 1
2^n≥1 Evident
_____________________Se aduna cele 2 rel;atii si se obtine
2^(n+1)≥n+2
Deci Pn=>Pn+1
