answer:1. **Proof**: Since the sequence {a_n} satisfies a_n=2a_{n-1}+2^n (for ngeq2 and ninmathbb{N}^*), we have: [ frac{a_n}{2^n}=frac{2a_{n-1}+2^n}{2^n}=frac{a_{n-1}}{2^{n-1}}+1. ] So the difference between consecutive terms is constant: [ frac{a_n}{2^n}-frac{a_{n-1}}{2^{n-1}}=1, ] Therefore, {frac{a_n}{2^n}} is an arithmetic sequence. 2. **General Term Formula**: We know that a_1=1, so frac{a_1}{2^1}=frac{1}{2}. From 1, we know that {frac{a_n}{2^n}} is an arithmetic sequence with a common difference of 1. Thus, [ frac{a_n}{2^n} = frac{1}{2} + (n - 1) = n - frac{1}{2}. ] Multiplying both sides by 2^n gives us the general term formula: [ a_n=(2n-1)2^{n-1}. ] 3. **Sum of the First n Terms, S_n**: Based on a_n=(2n-1)2^{n-1}, we can write S_n as [ S_n = 1cdot2^0 + 3cdot2^1 + 5cdot2^2 + ldots + (2n-1)2^{n-1}. ] Multiplying S_n by 2 gives [ 2S_n = 1cdot2^1 + 3cdot2^2 + 5cdot2^3 + ldots + (2n-1)2^n. ] Subtracting the first equation from the second yields [ -S_n = sum_{k=0}^{n-1} 2^{k+1} - (2n-1)2^n = 1 + (2sum_{k=1}^{n} 2^k) - (2n-1)2^n. ] Simplifying, we leverage the formula for the sum of a geometric series: [ -S_n = 1 + (2 cdot frac{2(1-2^{n-1})}{1-2}) - (2n-1)2^n = -3 + (3-2n)2^n, ] Therefore, the sum of the first n terms, S_n, is [ boxed{S_n = (2n-3)2^n + 3}. ]

answer:1. We are given that the set ( A = {a_1, a_2, a_3, a_4} ) consists of unknown elements, and their respective three-element subsets sum up to the elements in set ( B = {-1, 3, 5, 8} ). 2. Without loss of generality, assume ( a_1 < a_2 < a_3 < a_4 ). 3. Therefore, we can set up the following system of equations, corresponding to the sums given: [ begin{cases} a_1 + a_2 + a_3 = -1, a_1 + a_2 + a_4 = 3, a_1 + a_3 + a_4 = 5, a_2 + a_3 + a_4 = 8. end{cases} ] 4. To find ( a_1 + a_2 + a_3 + a_4 ), add all the equations together: [ (a_1 + a_2 + a_3) + (a_1 + a_2 + a_4) + (a_1 + a_3 + a_4) + (a_2 + a_3 + a_4) = -1 + 3 + 5 + 8. ] 5. Simplify the left-hand side: [ 3(a_1 + a_2 + a_3 + a_4) = 15. ] 6. Solve for ( a_1 + a_2 + a_3 + a_4 ): [ a_1 + a_2 + a_3 + a_4 = frac{15}{3} = 5. ] 7. Now use the sum ( a_1 + a_2 + a_3 + a_4 = 5 ) to solve for individual elements ( a_1, a_2, a_3, a_4 ) by substituting back into the original equations. 8. Subtract the equation ( a_1 + a_2 + a_3 = -1 ) from ( a_1 + a_2 + a_3 + a_4 = 5 ): [ a_4 = 5 - (-1) = 6. ] 9. Next, subtract ( a_1 + a_2 + a_4 = 3 ) from ( a_1 + a_2 + a_3 + a_4 = 5 ): [ a_3 = 5 - 3 = 2. ] 10. Then, subtract ( a_1 + a_3 + a_4 = 5 ) from ( a_1 + a_2 + a_3 + a_4 = 5 ): [ a_2 = 5 - 5 = 0. ] 11. Finally, subtract ( a_2 + a_3 + a_4 = 8 ) from ( a_1 + a_2 + a_3 + a_4 = 5 ): [ a_1 = 5 - 8 = -3. ] 12. Thus, the elements of set ( A ) are: [ A = {-3, 0, 2, 6}. ] Conclusion: [ boxed{{-3, 0, 2, 6}} ]

answer:Let's denote the length of the long sides as L and the length of the short sides as S. According to the problem, there are two long sides and two short sides, so the total length of the fence is: 2L + 2S = 640 feet We also know that 80 feet of fence need to be replaced, which is one of the short sides. Therefore, S = 80 feet. Now we can substitute S into the total length equation: 2L + 2(80) = 640 2L + 160 = 640 2L = 640 - 160 2L = 480 L = 480 / 2 L = 240 feet Now we have the lengths of the long sides (L = 240 feet) and the short sides (S = 80 feet). To find the ratio of the length of the long sides to the short sides, we divide the length of the long side by the length of the short side: Ratio (L:S) = L / S Ratio (L:S) = 240 / 80 Ratio (L:S) = 3 / 1 Therefore, the ratio of the length of the long sides to the short sides is boxed{3:1} .

answer:1. Substitute x = -2 into the polynomial expression x^3 - x^2 + x - 1. [ (-2)^3 - (-2)^2 + (-2) - 1 ] 2. Calculate each term: [ -8 - 4 - 2 - 1 = -15 ] 3. Sum all the terms to get the final result: [ boxed{-15} ]