1. We need to find a closed form expression for the sum $$\sum_{i=0}^{n-1} i(i+1)$$. 2. First, expand the term inside the summation: $$i(i+1) = i^2 + i$$ 3. So the sum can be written as: $$\sum_{i=0}^{n-1} (i^2 + i) = \sum_{i=0}^{n-1} i^2 + \sum_{i=0}^{n-1} i$$ 4. We know the formulas for these sums: - Sum of the first $m$ integers: $$\sum_{i=0}^{m} i = \frac{m(m+1)}{2}$$ - Sum of the first $m$ squares: $$\sum_{i=0}^{m} i^2 = \frac{m(m+1)(2m+1)}{6}$$ 5. Substitute $m = n-1$: $$\sum_{i=0}^{n-1} i = \frac{(n-1)n}{2}$$ $$\sum_{i=0}^{n-1} i^2 = \frac{(n-1)n(2n-1)}{6}$$ 6. Therefore: $$\sum_{i=0}^{n-1} i(i+1) = \frac{(n-1)n(2n-1)}{6} + \frac{(n-1)n}{2}$$ 7. Factor $\frac{(n-1)n}{6}$ from both terms: $$\sum_{i=0}^{n-1} i(i+1) = \frac{(n-1)n}{6} (2n - 1 + 3) = \frac{(n-1)n}{6} (2n + 2)$$ 8. Simplify: $$\frac{(n-1)n}{6} (2n + 2) = \frac{(n-1) n \cdot 2 (n + 1)}{6} = \frac{(n-1) n (n + 1)}{3}$$ **Final answer:** $$\sum_{i=0}^{n-1} i(i+1) = \frac{(n-1) n (n + 1)}{3}$$