[tex]P(n): 1\cdot 2+2\cdot 3+\ldots +n(n+1)=\dfrac{n(n+1)(n+2)}{3}\\\\P(1):1\cdot 2=\dfrac{1\cdot 2\cdot 3}{3}\\2=2(A) \\\\\text{Presupunem }P(k)-A\ \forall\ k\in \mathbb{Z}.\text{Demonstram ca }P(k+1)-A\\P(k):1\cdot 2+2\cdot 3+\ldots+ k(k+1)= \dfrac{k(k+1)(k+2)}{3},k\in \mathbb{Z} \\P(k+1):1\cdot 2+2\cdot 3+\ldots+ k(k+1)+(k+1)(k+2)=\\ \dfrac{(k+1)(k+2)(k+3)}{3}\\P(k+1):\dfrac{k(k+1)(k+2)}{3}+(k+1)(k+2)=\dfrac{(k+1)(k+2)(k+3)}{3}[/tex]
[tex]P(k+1):\dfrac{k(k+1)(k+2)+3(k+1)(k+2)}{3}=\dfrac{(k+1)(k+2)(k+3)}{3}\\P(k+1):\dfrac{(k+1)(k+2)(k+3)}{3}=\dfrac{(k+1)(k+2)(k+3)}{3}-A\\\Rightarrow P(k)-A\Rightarrow P(n)-A\\[/tex]
