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?

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.
