Răspuns:
Explicație pas cu pas:
[tex]a_{n} = \frac{1^2 + 2^2 + 3^2 + .... n^2}{n(1+2+3+ ... + n)} = \frac{\frac{n*(n+1)*(2n+1)}{6} }{n*\frac{n*(n+1)}{2} } = \\\\ =\frac{2*n*(n+1)*(2n+1)}{6*n*n*(n+1)} = \frac{2n+1}{3*n} = \frac{2*n}{3*n} + \frac{1}{3*n} = \frac{2}{3} +\frac{1}{3*n} \\\\ \lim_{n \to \infty} a_n = \lim_{n \to \infty} (\frac{2}{3} +\frac{1}{3*n}) = \lim_{n \to \infty} (\frac{2}{3}) + \lim_{n \to \infty} (\frac{1}{3*n}) = \frac{2}{3}[/tex]
