[tex]1^2+2^2+3^2+...+n^2 = \dfrac{n(n+1)(2n+1)}{6} \\ \\ \underline{\text{Demonstratie:}} \\ \\ \displaystyle \sum\limits_{k=1}^{n}\Big[(k+1)^3-k^3\Big] = \sum\limits_{k=1}^{n}(k^3+3k^2+3k+1-k^3)\\ \\ 2^3+3^3+...+(n+1)^3-1^3-2^3-...-n^3 =\sum\limits_{k=1}^{n}(3k^2+3k+1)\\ \\ (n+1)^3-1 = 3\sum\limits_{k=1}^n k^2+3\sum\limits_{k=1}^n k+n \\ \\(n+1)^3-1 = 3S+\dfrac{3n(n+1)}{2}+n \\ \\ 2(n+1)^3-2 = 6S+3n(n+1)+2n \\ \\ 6S = 2(n+1)^3-3n(n+1)-2n-2\\ \\ 6S = (n+1)\Big[2(n+1)^2-3n-2\Big] \\ \\ 6S =(n+1)(2n^2+n)[/tex]
[tex]6S =n(n+1)(2n+1) \\ \\ \Rightarrow 1^2+2^2+3^2+...+n^2 = \dfrac{n(n+1)(2n+1)}{6}[/tex]
[tex]\underline{\text{Rezolvare:}}[/tex]
[tex]1^2+2^2+3^2+...+100^2 = \dfrac{100(100+1)(2\cdot 100+1)}{6} \\ \\ 1^2+2^2+3^2+...+100^2 = \dfrac{100\cdot 101\cdot 201}{6}\\ \\ 1^2+2^2+3^2+...+100^2 = 338350[/tex]
