answer:Let's denote the time Malcolm walks uphill in the first route as ( x ) minutes. According to the problem, he walks for twice this amount of time along a path, which is ( 2x ) minutes, and finishes the journey in a third of the time it took to walk the first two stages, which is ( frac{1}{3}(x + 2x) = frac{1}{3}(3x) = x ) minutes. So, the total time for the first route is ( x + 2x + x = 4x ) minutes. For the second route, he walks along a flat path for 14 minutes, then finishes the walk in twice this amount of time, which is ( 2 times 14 = 28 ) minutes. So, the total time for the second route is ( 14 + 28 = 42 ) minutes. According to the problem, the second route was 18 minutes longer than the first. Therefore, we can write the equation: ( 4x + 18 = 42 ) Now, let's solve for ( x ): ( 4x = 42 - 18 ) ( 4x = 24 ) ( x = frac{24}{4} ) ( x = 6 ) Malcolm walked uphill for boxed{6} minutes in the first route.

answer:1. Given the sequence {a_n} defined by the recurrence relation: [ a_1 = 1, quad a_{n+1} = frac{n a_n + 2(n+1)^2}{n+2} ] 2. Starting with the recurrence relation, multiply both sides by (n+2) to clear the denominator: [ (n+2)a_{n+1} = n a_n + 2(n+1)^2 ] 3. Rewrite 2(n+1)^2 in a more convenient form: [ 2(n+1)^2 = 2(n^2 + 2n + 1) = 2n^2 + 4n + 2 ] 4. Next, multiply the left side by (n+1) to both sides of the equation: [ (n+1)(n+2)a_{n+1} = (n+1)[n a_n + 2(n^2 + 2n + 1)] ] 5. Introduce a new sequence b_n = n(n+1)a_n. Compute the first term: [ b_1 = 1 times 2 times a_1 = 2 ] 6. Now express b_{n+1} in terms of b_n: [ b_{n+1} = (n+1)(n+2)a_{n+1} ] 7. Substitute a_{n+1} with the recurrence relation: [ b_{n+1} = (n+1)[n a_n + 2(n+1)^2] = (n+1)n a_n + 2(n+1)^3 ] 8. Observe: [ (n+1)n a_n = b_n ] 9. Therefore: [ b_{n+1} = b_n + 2(n+1)^3 ] 10. Now, expand for b_n: [ b_n = b_1 + sum_{i=1}^{n-1} 2(i+1)^3 ] 11. Compute the sum: [ sum_{i=1}^{n-1} (i+1)^3 = 1^3 + 2^3 + cdots + n^3 ] 12. Using the sum of cubes formula: [ 1^3 + 2^3 + cdots + n^3 = left(frac{n(n+1)}{2}right)^2 ] 13. Then, [ b_n = b_1 + 2left(frac{n(n+1)}{2}right)^2 = 2 + 2left(frac{n(n+1)}{2}right)^2 ] 14. Simplify: [ b_n = frac{1}{2} n^2(n+1)^2 ] 15. Finally: [ a_n = frac{b_n}{n(n+1)} = frac{frac{1}{2} n^2 (n+1)^2}{n(n+1)} = frac{1}{2} n(n+1) ] 16. The general term of the sequence is: [ a_n = frac{1}{2} n(n+1) ] 17. Summing for S_n: [ S_n = frac{1}{2}left(sum_{k=1}^n k^2 + sum_{k=1}^n kright) ] 18. Use known sums: [ sum_{k=1}^n k^2 = frac{n(n+1)(2n+1)}{6}, quad sum_{k=1}^n k = frac{n(n+1)}{2} ] 19. Combine these: [ S_n = frac{1}{2} left[ frac{n(n+1)(2n+1)}{6} + frac{n(n+1)}{2} right] ] 20. Simplify: [ S_n = frac{1}{2} left[ frac{n(n+1)(2n+1) + 3n(n+1)}{6} right] = frac{1}{2} left[ frac{n(n+1)(2n+1 + 3)}{6} right] = frac{1}{2} left[ frac{n(n+1)(n+2)}{6} right] ] 21. The sum: [ S_n = frac{1}{6} n(n+1)(n+2) ] Conclusion: [ boxed{frac{1}{6} n(n+1)(n+2)} ]

answer:Since the universal set U={0,1,2,3,4}, and since the set A={1,2}, therefore, C_UA={0,3,4}. Hence, the correct option is boxed{C}.

answer:First, calculate the taxable income: [ text{Taxable income} = 500x - 2000. ] Deduct this from the original income: [ text{Tax} = frac{x}{100} cdot (500x - 2000). ] Calculate the tax: [ text{Tax} = 5x^2 - frac{20x}{100}x = 5x^2 - 0.2x^2 = 4.8x^2. ] Now determine the take-home pay: [ text{Take-home pay} = 500x - 4.8x^2. ] Completing the square to find the maximum take-home pay: [ 500x - 4.8x^2 = -4.8(x^2 - frac{500}{4.8}x) = -4.8left(x^2 - frac{500}{4.8}x + left(frac{250}{4.8}right)^2right) + 4.8left(frac{250}{4.8}right)^2 = -4.8left(x - 52.08right)^2 + 4.8 cdot 569.44. ] The maximum take-home pay occurs when x = 52.08, approximately corresponding to an income of boxed{26040} dollars.