1. The problem is to analyze the summation $$\sum_{i=1}^n iL(2)^{x+2} = (n-2)^{x+2} A$$ with the condition $$n > 0$$. 2. First, understand the summation notation: $$\sum_{i=1}^n iL(2)^{x+2}$$ means adding the terms $$iL(2)^{x+2}$$ for $$i$$ from 1 to $$n$$. 3. Since $$L(2)^{x+2}$$ does not depend on $$i$$, it can be factored out: $$\sum_{i=1}^n iL(2)^{x+2} = L(2)^{x+2} \sum_{i=1}^n i$$ 4. The sum of the first $$n$$ natural numbers is: $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ 5. Substitute this back: $$L(2)^{x+2} \cdot \frac{n(n+1)}{2} = (n-2)^{x+2} A$$ 6. This equation relates $$n$$, $$x$$, $$L$$, and $$A$$. To solve for $$A$$: $$A = \frac{L(2)^{x+2} \cdot \frac{n(n+1)}{2}}{(n-2)^{x+2}} = \frac{L(2)^{x+2} n(n+1)}{2 (n-2)^{x+2}}$$ 7. Important notes: - $$n > 0$$ and $$n \neq 2$$ to avoid division by zero. - $$L$$ and $$A$$ are constants or parameters depending on context. Final answer: $$A = \frac{L(2)^{x+2} n(n+1)}{2 (n-2)^{x+2}}$$