P(1): 4*1-3=1*1 <=> 1=1 (A)
Presupunem ca P(k) e adv. si dem.P(k+1)
P(k):1+5+9+...(4k-3)=k(2k-1)
P(k+1): 1+5+9+...+(4k-3)+4(k+1)-3=(k+1)[2(k+1)-1]
k(2k-1)+4k+1=(k+1)(2k+1)
2k²+3k+1=2k²+3k+1 (A)
si se reduc, dand 0=0 (A)
Conform principiului ind.matematice =>P(n) e adv.
