P(n): 2^n>n^2
1)P(5):2^5>5^2<=>32>25(Adevarat)
2)Presupunem P(n) (A) si demonstram P(n+1)(A)
Aratam ca P(n+1): 2^{n+1}>(n+1)^2
2^n>n^2 si inmultim cu 2=>
2^{n+1}>2n^2>(n+1)^2
Aratam ca 2n^2>(n+1)^2.
2n^2>n^2+2n+1
n^2-2n-1>0<=>(n-1)^2-2>0 adevarata pentru n≥5