n = mumarul tremenilor !
[tex]Exist\u{a}\;o\;formul\u{a}\;de\;forma:\\
\;1^2+2^2+3^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\;;\\
Si\;aplic\u{a}m\;dup\u{a}\;nevoi\;ideea:\\
ex.\;1^2+3^2+5^2+...+(2n-1)^2=\\
=1^1+2^2+3^2+...+n^2 - (2^2+4^2+...+(2n)^2=\\
=\frac{n(n+1)(2n+1)}{6}-2^2(\underbrace{1^2+2^2+...+n^2}_{formula})=...
[/tex]