find the limit of a sequnce I need to find the limit:
$\mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {{{(a + {1 \over n})}^2} + {{(a + {2 \over n})}^2} + ... + {{(a + {{n - 1} \over n})}^2}} \right]$  
any ideas here? I've tried to use "squeeze theorem" but with no luck..  

Let $u_n={1 \over n}\left[ {{{(a + {1 \over n})}^2} + {{(a + {2 \over n})}^2} + ... + {{(a + {{n - 1} \over n})}^2}} \right]$.
To elaborate on Daniel Fischer’answer : the Riemann sum approach yields
$u_n=\sum_{k=1}^{n-1} \frac{1}{n}f(\frac{k}{n})$ where $f(x)=(a+x)^2$, so $(u_n)$ converges to $\int_0^1 f(x)dx=\frac{(a+1)^3-a^3}{3}=\frac{3a^2+3a+1}{3}$.
Also, the expansion approach yields
$$
\begin{array}{lcl}
u_n &=& \frac{1}{n}. \sum_{k=1}^{n-1} (a+\frac{k}{n})^2 \\
  &=& \frac{1}{n}. \sum_{k=1}^{n-1} \big(a^2+2\frac{ak}{n}+\frac{k^2}{n^2}\big) \\
  &=& (\frac{n-1}{n})a^2+\frac{2a}{n^2}\big(\sum_{k=1}^{n-1}k\big)
  +\frac{1}{n^3}\big(\sum_{k=1}^{n-1}k^2\big) \\
  &=& (\frac{n-1}{n})a^2+\frac{a(n-1)}{n}
  +\frac{1}{n^3}\big(\frac{n(n-1)(2n-1)}{6}\big)
\end{array}
$$
And we see again that $(u_n)$ converges to $\frac{3a^2+3a+1}{3}$.
