1. The problem is to evaluate the sum $$\sum_{i=1}^{n} i^2 (5-1)$$. 2. Simplify the term inside the sum: $$5-1 = 4$$, so the sum becomes $$\sum_{i=1}^{n} 4 i^2$$. 3. Factor out the constant 4 from the sum: $$4 \sum_{i=1}^{n} i^2$$. 4. Use the formula for the sum of squares: $$\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$$. 5. Substitute the formula into the expression: $$4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{4 n(n+1)(2n+1)}{6}$$. 6. Simplify the fraction: $$\frac{4}{6} = \frac{2}{3}$$, so the final answer is $$\frac{2 n(n+1)(2n+1)}{3}$$. Therefore, $$\sum_{i=1}^{n} i^2 (5-1) = \frac{2 n(n+1)(2n+1)}{3}$$.