answer:(1) From f(x)=(x^{2}+ax-2a-3)e^{x}, we find the derivative f'(x)=(2x+a)e^{x}+e^{x}(x^{2}+ax-2a-3)=[x^{2}+(a+2)x-a-3]e^{x}. Since f(x) reaches an extreme value at x=2, we have f'(2)=0, which leads to 4+(a+2) times 2 - a - 3 = 0. Solving for a, we get a=-5. So, the value of a is boxed{a=-5}. Then, f(x)=(x^{2}-5x+7)e^{x}. (2) From f(x)=(x^{2}-5x+7)e^{x}, we find the derivative f'(x)=(x^{2}-3x+2)e^{x}=(x-2)(x-1)e^{x}. Solving f'(x)=0, we get x=1 or x=2. So, f(x) is increasing on (-infty, 1) and (2, +infty), and decreasing on (1, 2). When x=2, f(x) reaches the minimum value, which is f(2)=e^{2}. Now, let's evaluate f(x) at the endpoints of the interval [frac{3}{2}, 3]. We have f(frac{3}{2})=frac{7}{4}e^{frac{3}{2}} and f(3)=e^{3}. Since f(3)-f(frac{3}{2})=e^{3}-e^{frac{3}{2}}=frac{1}{4}e^{frac{3}{2}}(4e^{sqrt{e}}-7)>0, we have f(3) > f(frac{3}{2}). Therefore, the maximum value of f(x) on the interval [frac{3}{2}, 3] is boxed{f(3)=e^{3}}, and the minimum value of f(x) on the interval [frac{3}{2}, 3] is boxed{f(2)=e^{2}}.

answer:This problem primarily focuses on determining the range of a parameter based on the relationships among the terms in the sequence. The key to solving this problem is to derive a relationship involving a_n from the given conditions. First, from S_n + S_{n-1} = 4n^2 for n geq 2, n in mathbb{N^+}, we can deduce that S_{n+1} + S_{n} = 4(n+1)^2. Subtracting the two expressions yields S_{n+1} - S_{n-1} = 8n+4. This can be simplified to a_{n+1} + a_n = 8n + 4. Then, we have a_{n+2} + a_{n+1} = 8n + 12. Subtracting the two expressions again yields a_{n+2} - a_n = 8. Setting n=2, we have a_2 + 2a_1 = 16, which gives a_2 = 16 - 2a. For n=3, we have a_3 = 2a + 4. And for n=4, we have a_4 = -2a + 24. Given that a_n < a_{n+1} for any n in mathbb{N^+}, we only need to ensure that a_1 < a_2 < a_3 < a_4. This leads to the inequality a < 16 - 2a < 4 + 2a < 24 - 2a. Solving this inequality, we get 3 < a < 5. Thus, the answer is boxed{(3, 5)}.

answer:1. **Given Data and Initial Setup:** - In triangle ABC, I is the incenter. - AI intersects BC at D. - AI = 3, ID = 2, and BI^2 + CI^2 = 64. - We need to find BC^2. 2. **Using the Law of Sines in triangle AIB:** [ AI = frac{AB sin frac{B}{2}}{sin left(frac{A}{2} + frac{B}{2}right)} ] Given AI = 3, we have: [ 3 = 4R sin frac{B}{2} sin frac{C}{2} ] 3. **Using the Law of Sines in triangle BID:** [ ID = frac{BI sin frac{B}{2}}{sin left(B + frac{A}{2}right)} ] Given ID = 2, we have: [ 2 = frac{4R sin frac{A}{2} sin frac{B}{2} sin frac{C}{2}}{sin left(B + frac{A}{2}right)} ] 4. **Combining the Two Equations:** [ frac{3 sin frac{A}{2}}{sin left(B + frac{A}{2}right)} = 2 implies 2 sin left(B + frac{A}{2}right) = 3 sin frac{A}{2} ] 5. **Using Sum-to-Product Identities:** [ sin frac{A}{2} = 2 left[sin left(B + frac{A}{2}right) - sin frac{A}{2}right] ] [ = 4 cos left(frac{A + B}{2}right) sin frac{B}{2} ] [ = 4 sin frac{C}{2} sin frac{B}{2} quad (diamondsuit) ] 6. **Simplifying the Third Condition:** [ BI^2 + CI^2 = 64 ] [ 16R^2 sin^2 frac{A}{2} left(sin^2 frac{B}{2} + sin^2 frac{C}{2}right) = 64 ] [ 256R^2 sin^2 frac{B}{2} sin^2 frac{C}{2} left(sin^2 frac{B}{2} + sin^2 frac{C}{2}right) = 64 ] [ sin^2 frac{B}{2} + sin^2 frac{C}{2} = frac{4}{9} ] 7. **Using the Identity:** [ cos left(frac{B}{2} + frac{C}{2}right) = 4 sin frac{C}{2} sin frac{B}{2} ] [ cos frac{C}{2} cos frac{B}{2} - sin frac{C}{2} sin frac{B}{2} = 4 sin frac{C}{2} sin frac{B}{2} ] [ cos frac{C}{2} cos frac{B}{2} = 5 sin frac{C}{2} sin frac{B}{2} ] [ sqrt{left(1 - sin^2 frac{C}{2}right) left(1 - sin^2 frac{B}{2}right)} = 5 sin frac{C}{2} sin frac{B}{2} ] [ sqrt{frac{5}{9} + sin^2 frac{B}{2} sin^2 frac{C}{2}} = 5 sin frac{C}{2} sin frac{B}{2} ] 8. **Solving for sin frac{B}{2} sin frac{C}{2}:** [ sin frac{B}{2} sin frac{C}{2} = frac{sqrt{30}}{36} ] [ sin frac{A}{2} = frac{sqrt{30}}{9} ] 9. **Finding sin A:** [ sin A = 2 cdot frac{sqrt{30}}{9} cdot frac{sqrt{51}}{9} = frac{2sqrt{170}}{27} ] 10. **Finding R:** [ R = frac{3}{4 sin frac{B}{2} sin frac{C}{2}} = frac{9sqrt{30}}{10} ] 11. **Calculating BC^2:** [ BC^2 = left(frac{9sqrt{30}}{10} cdot frac{2sqrt{170}}{27} cdot 2right)^2 = boxed{frac{272}{3}} ]

answer:1. Start by adding 6 to both sides of the inequality: [ 1.5n - 6 < 7.5 implies 1.5n < 13.5 ] 2. Next, divide both sides by 1.5 to isolate n: [ n < 9 ] 3. The positive integers less than 9 are 1, 2, 3, 4, 5, 6, 7, 8. Their sum is: [ 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36 ] Thus, the sum of all positive integers n satisfying the inequality is boxed{36}.