1. **State the problem:** Find the limit as $n \to \infty$ of $$\lim_{n \to \infty} \frac{1}{n} \left((a + \frac{1}{n})^2 + (a + \frac{2}{n})^2 + \cdots + (a + \frac{n-1}{n})^2\right).$$ 2. **Rewrite the sum:** The sum inside is $$\sum_{k=1}^{n-1} \left(a + \frac{k}{n}\right)^2 = \sum_{k=1}^{n-1} \left(a^2 + 2a \frac{k}{n} + \frac{k^2}{n^2}\right).$$ 3. **Distribute the sum:** $$\sum_{k=1}^{n-1} a^2 + \sum_{k=1}^{n-1} 2a \frac{k}{n} + \sum_{k=1}^{n-1} \frac{k^2}{n^2} = a^2 (n-1) + 2a \frac{1}{n} \sum_{k=1}^{n-1} k + \frac{1}{n^2} \sum_{k=1}^{n-1} k^2.$$ 4. **Use formulas for sums:** $$\sum_{k=1}^{m} k = \frac{m(m+1)}{2}, \quad \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6}.$$ Apply for $m = n-1$: $$\sum_{k=1}^{n-1} k = \frac{(n-1)n}{2}, \quad \sum_{k=1}^{n-1} k^2 = \frac{(n-1)n(2n-1)}{6}.$$ 5. **Substitute back:** $$a^2 (n-1) + 2a \frac{1}{n} \cdot \frac{(n-1)n}{2} + \frac{1}{n^2} \cdot \frac{(n-1)n(2n-1)}{6} = a^2 (n-1) + a (n-1) + \frac{(n-1)(2n-1)}{6n}.$$ 6. **Divide entire sum by $n$ as in the original limit:** $$\frac{1}{n} \left(a^2 (n-1) + a (n-1) + \frac{(n-1)(2n-1)}{6n}\right) = \frac{a^2 (n-1)}{n} + \frac{a (n-1)}{n} + \frac{(n-1)(2n-1)}{6n^2}.$$ 7. **Simplify each term and take the limit as $n \to \infty$:** $$\lim_{n \to \infty} \frac{a^2 (n-1)}{n} = a^2, \quad \lim_{n \to \infty} \frac{a (n-1)}{n} = a,$$ $$\lim_{n \to \infty} \frac{(n-1)(2n-1)}{6n^2} = \lim_{n \to \infty} \frac{2n^2 - 3n + 1}{6n^2} = \frac{2}{6} = \frac{1}{3}.$$ 8. **Add the limits:** $$a^2 + a + \frac{1}{3}.$$ **Final answer:** $$\boxed{a^2 + a + \frac{1}{3}}.$$