2. Să se demonstreze egalitatea: \(1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6},\forall n\in \mathbf N^*\)
Soluţie. \begin{align} &\cssId{Step1}{I. Etapa\; de\; verificare:}\\ &\cssId{Step2}{ P(1): 1^2=\frac{1(1+1)(2\cdot 1+1)}{6} \Leftrightarrow 1=\frac{6}{6} \Leftrightarrow 1=1\;"A"}\\ &\cssId{Step3}{II. Etapa\; de\; demonstraţie:\; Presupunem\; adevărată}\\ &\cssId{Step4}{P(k): 1^2+2^2+3^2+...+k^2=\frac{k(k+1)(2k+1)}{6}}\\ &\cssId{Step5}{şi\; demonstrăm\; P(k+1): 1^2+2^2+3^2+...+k^2+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}?}\\ &\cssId{Step6}{P(k+1):\underbrace{1^2+2^2+3^2+...+k^2}_{P(k)} +(k+1)^2=\frac{k(k+1)(2k+1)}{6}+^{6)}(k+1)^2=}\\ &\cssId{Step7}{=\frac{(k+1)[k(2k+1)+6(k+1)]}{6}=\frac{(k+1)(2k^2+7k+6)}{6}**}\\ &\cssId{Step8}{Ecuaţia ;2k^2+7k+6=0\; are\; rădăcinile \;k_1=-2,\;k_2=-\frac{3}{2} \Rightarrow 2k^2+7k+6=2(k+\frac{3}{2} )(k+2)=(2k+3)(k+2)}\\ &\cssId{Step9}{Revenim\; **\;\frac{(k+1)(2k^2+7k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}\Rightarrow P(k+1) "A"}\\ &\cssId{Step10}{Din\; I.\; şi\; II.\;\Rightarrow P(n)"A", \forall n \in \mathbf N^*}\\ \end{align}