Calculus limit homework problem $$ \lim_{n→\infty} \frac1 n \left(\left(a + \frac 1 n\right)^2 + \left(a + \frac 2 n\right)^2 + ... + \left(a + \frac{n-1}{n}\right)^2\right)$$ 
$$ \text{hint: }\  1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6} $$
I can't figure out how to find this problem's limit. Does anybody have ideas?
 A: \begin{align*}
 & \; \; \; \lim_{n\to \infty}\frac 1 n \left(  \left(a + \frac 1 n \right )^2 + \dots +\left(a + \frac {n-1} n \right )^2 \right ) \\ 
 &= \lim_{n\to \infty} \frac 1 n \left( (n-1)a^2  + 2 a \left( \frac 1 n + \dots  + \frac{n-1}{n} \right ) + \left( \frac {1^n}{n^2} + \dots + \frac{(n-1)^2}{n^2}\right )\right )\\ 
 &= \lim_{n\to \infty}\frac{(n-1)a^2}{n} + 2a \frac{n(n-1)}{ 2 n^2} + \frac{n(n-1)(2n-1)}{6n^3}\\ 
 &= a^2 + a + \frac 1 3 \\
\end{align*}
Also it can be evaluated via converting it into Integral noticing that as $n\to \infty $  as
$$\lim_{n\to \infty}\frac 1 n \left(  \left(a + \frac 1 n \right )^2 + \dots +\left(a + \frac {n-1} n \right )^2 \right )= \lim_{n\to \infty } \frac 1 n  \sum_{i=0}^{n-1} \left( a + \frac i n \right )^2= \int_0^1 \left( a + x \right )^2 dx = a^2 + a + \frac 1 3 $$
A: Ignore the limit for now. We will find the expression first, and then evaluate the limit.
$$\frac{1}{n}\left((a^2+2a/n+1/n^2)+(a^2+4a/n+4/n^2) + \ldots + (a^2 + (k) 2a/n  + k^2/n^2) + \ldots + (a^2+(n-1)2a/n) + (n-1)^2/n^2)\right).$$
This simplifies to $$\frac{1}{n}((n-1)a^2 + 2a/n (1 + 2 + \ldots + n - 1) + 1/n^2 (1^2 + 2^2 + \ldots + (n-1)^2).$$
Use $$1 + 2 + \ldots + n - 1 = n(n - 1)/2$$ and the other identity given to you, and you should be set.
A: The main observation here is that
\begin{align}
\frac{1}{n}\left(\left(a + \frac 1 n\right)^2 + \left(a + \frac 2 n\right)^2 + ... + \left(a + \frac{n-1}{n}\right)^2\right)&= a^2+2a\sum_{i=1}^{n-1}i+\frac{1}{n^3}\sum_{i=1}^{n-1}i^2.
\end{align}
Then, the third term in the right hand side of the above equality is given in the hint, and the second term is simply an arithmetic sum (http://en.wikipedia.org/wiki/Arithmetic_progression). 
