Explicație pas cu pas:
1.Metoda verificarii:
n=1;P1:1²=1(1+1)(2+1)/6⇒1=1 adev.
2. n=k
Pk: 1²+2²+3²+...+k²=k(k+1)(2k+1)/6
3.n=k+1
Pk+1: 1²+2²+3²+...+(k+1)²=(k+1)(k+2)(2k+3)/6 ?-trebuie sa o demonstram
1²+2²+3²+...k²+(k+1)²=k(k+1)(2k+1)/6+(k+1)²=[k(k+1)(2k+1)+6(k+1)²]/6=
=[(k+1)(k(2k+1)+6(k+1)]/6=[(k+1)(2k²+k+6k+6)]/6=[(k+1)(2k²+7k+6)]/6=
=[(k+1)(k+2)(2k+3)]/6 deci este adevarat si pentru n=k+1.
Raspuns: Pn adevarat ∀ n∈ N*.
Bafta!
