1. **Problem:** Find the limit of the sequence \(\lim_{n \to \infty} \frac{1 + 4 + 7 + \dots + (3n - 2)}{\sqrt{5n^4 + n + 1}}\). 2. **Formula and rules:** The numerator is an arithmetic series with first term \(a_1 = 1\), common difference \(d = 3\), and \(n\) terms. The sum of the first \(n\) terms is \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\). 3. **Calculate numerator:** \[S_n = \frac{n}{2}(2 \cdot 1 + (n-1) \cdot 3) = \frac{n}{2}(2 + 3n - 3) = \frac{n}{2}(3n - 1) = \frac{3n^2 - n}{2}\] 4. **Calculate denominator:** \[\sqrt{5n^4 + n + 1} = \sqrt{5n^4 \left(1 + \frac{n}{5n^4} + \frac{1}{5n^4}\right)} = n^2 \sqrt{5} \sqrt{1 + \frac{1}{5n^3} + \frac{1}{5n^4}}\] As \(n \to \infty\), the terms inside the square root approach 1, so denominator \(\sim n^2 \sqrt{5}\). 5. **Form the limit:** \[\lim_{n \to \infty} \frac{\frac{3n^2 - n}{2}}{n^2 \sqrt{5}} = \lim_{n \to \infty} \frac{3n^2 - n}{2 n^2 \sqrt{5}} = \lim_{n \to \infty} \frac{3 - \frac{1}{n}}{2 \sqrt{5}} = \frac{3}{2 \sqrt{5}}\] 6. **Simplify the final answer:** \[\frac{3}{2 \sqrt{5}} = \frac{3 \sqrt{5}}{2 \cdot 5} = \frac{3 \sqrt{5}}{10}\] **Final answer:** \(\boxed{\frac{3 \sqrt{5}}{10}}\).