Răspuns :

p(n):1+5+....+(4n-3)=n(2n-1)

Etapa I
n=1=>4*1-3=1*(2*1-1)
1=1 (A)

Etapa II p(k)->p(k+1), k≥0 fixat
p(k): 1+5+.....+(4k-3)=k(2k-1) (A)
p(k+1): 1+5+....+(4k-3)+(4k+1)=(k+1)(2k+1)
p(k)+(4k+1)=(k+1)(2k+1)
k(2k-1)+(4k+1)=(k+1)(2k+1)
2k²-k+4k+1=2k²+k+2k+1
2k²+3k+1=2k²+3k+1 (A)
=>p(k)->p(k+1) (A)

I, II=> v(p(n))=1 pt ∀n≥0