1. **State the problem:** We need to find the sum $$\sum_{i=1}^{20} \left( x^2 \sqrt{\frac{9x^4}{x^8}} \sqrt[i]{4} + \sqrt{\frac{125}{5}}^{2 \log_5 5} + \log_3 243 \right)$$. 2. **Simplify each term inside the sum:** - Simplify $$\sqrt{\frac{9x^4}{x^8}}$$: $$\frac{9x^4}{x^8} = 9x^{4-8} = 9x^{-4}$$ $$\sqrt{9x^{-4}} = \sqrt{9} \cdot \sqrt{x^{-4}} = 3 \cdot x^{-2}$$ - So the first part inside the sum becomes: $$x^2 \cdot 3x^{-2} \cdot \sqrt[i]{4} = 3 \cdot x^{2-2} \cdot 4^{\frac{1}{i}} = 3 \cdot 4^{\frac{1}{i}}$$ - Simplify $$\sqrt{\frac{125}{5}}^{2 \log_5 5}$$: $$\frac{125}{5} = 25$$ $$\sqrt{25} = 5$$ Since $$\log_5 5 = 1$$, then: $$5^{2 \cdot 1} = 5^2 = 25$$ - Simplify $$\log_3 243$$: Since $$243 = 3^5$$, then: $$\log_3 243 = 5$$ 3. **Rewrite the sum:** $$\sum_{i=1}^{20} \left( 3 \cdot 4^{\frac{1}{i}} + 25 + 5 \right) = \sum_{i=1}^{20} \left( 3 \cdot 4^{\frac{1}{i}} + 30 \right)$$ 4. **Separate the sum:** $$\sum_{i=1}^{20} 3 \cdot 4^{\frac{1}{i}} + \sum_{i=1}^{20} 30 = 3 \sum_{i=1}^{20} 4^{\frac{1}{i}} + 30 \cdot 20$$ 5. **Calculate the constant part:** $$30 \cdot 20 = 600$$ 6. **Final expression:** $$3 \sum_{i=1}^{20} 4^{\frac{1}{i}} + 600$$ This is the simplified form of the sum. The exact numeric value depends on evaluating $$\sum_{i=1}^{20} 4^{\frac{1}{i}}$$ which is a sum of 20 terms. **Answer:** $$\boxed{3 \sum_{i=1}^{20} 4^{\frac{1}{i}} + 600}$$