1. **State the problem:** Find the value of the summation $$\sum_{i=1}^{50} (2i + 1)$$. 2. **Recall the formula:** We know the formula for the sum of the first $n$ integers is $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$. 3. **Rewrite the summation:** The summation can be split as $$\sum_{i=1}^{50} (2i + 1) = \sum_{i=1}^{50} 2i + \sum_{i=1}^{50} 1$$ 4. **Calculate each part:** - $$\sum_{i=1}^{50} 2i = 2 \sum_{i=1}^{50} i = 2 \times \frac{50 \times 51}{2}$$ - $$\sum_{i=1}^{50} 1 = 50$$ (since adding 1 fifty times) 5. **Simplify:** $$2 \times \frac{50 \times 51}{2} = \cancel{2} \times \frac{50 \times 51}{\cancel{2}} = 50 \times 51 = 2550$$ 6. **Add the parts:** $$2550 + 50 = 2600$$ **Final answer:** $$\sum_{i=1}^{50} (2i + 1) = 2600$$