1. **State the problem:** We are given two summations involving terms $y_i$: $$\sum_{i=1}^{4p^2-p} y_i = 4p - 1 - x$$ $$\sum_{i=1}^{4p^2-p} i \cdot y_i = (4p - 1)p$$ We want to analyze or solve for $y_i$ or related expressions based on these equations. 2. **Understand the summations:** The first summation is the sum of all $y_i$ from $i=1$ to $4p^2 - p$. The second summation is the weighted sum of $y_i$ with weights $i$. 3. **Key formulas and rules:** These are linear equations in terms of $y_i$. If we consider $y_i$ as unknowns, these two equations alone are insufficient to solve for all $y_i$ individually because there are $4p^2 - p$ unknowns but only two equations. 4. **Possible approach:** If $y_i$ follow a pattern or are related linearly, for example $y_i = a + b i$, then we can use these sums to find $a$ and $b$. 5. **Sum formulas:** Recall the formulas for sums of integers: $$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$ $$\sum_{i=1}^n 1 = n$$ 6. **Assume $y_i = a + b i$ and substitute:** $$\sum_{i=1}^n y_i = \sum_{i=1}^n (a + b i) = a n + b \frac{n(n+1)}{2}$$ $$\sum_{i=1}^n i y_i = \sum_{i=1}^n i (a + b i) = a \frac{n(n+1)}{2} + b \frac{n(n+1)(2n+1)}{6}$$ where $n = 4p^2 - p$. 7. **Set up the system:** $$a n + b \frac{n(n+1)}{2} = 4p - 1 - x$$ $$a \frac{n(n+1)}{2} + b \frac{n(n+1)(2n+1)}{6} = (4p - 1)p$$ 8. **Solve for $a$ and $b$:** Multiply the first equation by $\frac{n+1}{2}$ and subtract from the second to eliminate $a$: $$a \frac{n(n+1)}{2} + b \frac{n(n+1)(2n+1)}{6} - \left(a n \frac{n+1}{2} + b \frac{n(n+1)^2}{4}\right) = (4p - 1)p - (4p - 1 - x) \frac{n+1}{2}$$ Simplify the $a$ terms: $$a \frac{n(n+1)}{2} - a n \frac{n+1}{2} = 0$$ So the $a$ terms cancel out. Simplify the $b$ terms: $$b \frac{n(n+1)(2n+1)}{6} - b \frac{n(n+1)^2}{4} = b n (n+1) \left(\frac{2n+1}{6} - \frac{n+1}{4}\right)$$ Calculate inside the parentheses: $$\frac{2n+1}{6} - \frac{n+1}{4} = \frac{4(2n+1) - 6(n+1)}{24} = \frac{8n + 4 - 6n - 6}{24} = \frac{2n - 2}{24} = \frac{n - 1}{12}$$ So the $b$ term is: $$b n (n+1) \frac{n - 1}{12}$$ The right side is: $$(4p - 1)p - (4p - 1 - x) \frac{n+1}{2}$$ 9. **Solve for $b$:** $$b = \frac{12 \left[(4p - 1)p - (4p - 1 - x) \frac{n+1}{2}\right]}{n (n+1) (n - 1)}$$ 10. **Solve for $a$ using first equation:** $$a = \frac{4p - 1 - x - b \frac{n(n+1)}{2}}{n}$$ **Final answer:** $$a = \frac{4p - 1 - x - b \frac{n(n+1)}{2}}{n}, \quad b = \frac{12 \left[(4p - 1)p - (4p - 1 - x) \frac{n+1}{2}\right]}{n (n+1) (n - 1)}, \quad n = 4p^2 - p$$