answer:Let's analyze each conclusion step by step, following the definition of a "doubling point" and the given conditions. ① For points Q_{1}(2,8) and Q_{2}(-3,-2) to be "doubling points" of point P_{1}(2,0), we check the condition 2(x_{1}+x_{2})=y_{1}+y_{2} for each: - For Q_{1}(2,8): 2(2+2) = 8 + 0 Rightarrow 8 = 8, which satisfies the condition. - For Q_{2}(-3,-2): 2(2-3) = -2 + 0 Rightarrow -2 = -2, which also satisfies the condition. Thus, conclusion ① is correct. ② For a point A on the line y=x+2 to be a "doubling point" of P_{1}(2,0), we substitute y with x+2 in the "doubling point" condition: 2(x_{1}+x) = y_{1} + (x+2) Rightarrow 2(2+x) = 0 + (x+2) Rightarrow 4 + 2x = x + 2 Solving for x gives x = -2. Thus, A(-2, 0), which means conclusion ② is correct. ③ For a point on the parabola y=x^{2}-2x-3 to be a "doubling point" of P_{1}(2,0), we substitute y with x^{2}-2x-3: 2(x+2) = 0 + (x^{2}-2x-3) Rightarrow 4 + 2x = x^{2}-2x-3 Rearranging gives x^{2}-4x-7=0. The discriminant Delta = (-4)^{2} - 4(1)(-7) = 16 + 28 = 44 > 0, indicating two real solutions. Therefore, conclusion ③ is correct. ④ For any "doubling point" B(x,y) of P_{1}(2,0), we have 2(x+2)=y+0 Rightarrow y=2(x+2). The distance P_{1}B is given by: P_{1}B = sqrt{(x-2)^{2}+y^{2}} = sqrt{(x-2)^{2}+4(x+2)^{2}} = sqrt{5x^{2}+frac{48x}{5}+frac{64}{5}} Minimizing P_{1}B with respect to x gives x = -frac{6}{5}, leading to the minimum distance P_{1}B = frac{8sqrt{5}}{5}. Thus, conclusion ④ is correct. Since all conclusions ①, ②, ③, and ④ are correct, the number of correct conclusions is boxed{D}.

answer:Given that the sequence of positive terms {a_n} satisfies: a_n^2 + (1 - n)a_n - n = 0, we can factorize the quadratic equation to: (a_n - n)(a_n + 1) = 0, which gives a_n = n. Thus, b_n = frac{1}{(n + 1)a_n} = frac{1}{n(n + 1)} = frac{1}{n} - frac{1}{n + 1}. Now let's find T_n: T_n = b_1 + b_2 + ... + b_n = (1 - frac{1}{2}) + (frac{1}{2} - frac{1}{3}) + ... + (frac{1}{n} - frac{1}{n + 1}) = 1 - frac{1}{n + 1} Therefore, T_{2016} = 1 - frac{1}{2017} = boxed{frac{2016}{2017}}. To solve this problem, we first find the general term a_n of the sequence {a_n} by factoring the quadratic equation. Then, we simplify b_n using the general term a_n and calculate the sum of the first n terms of the sequence {b_n} using the method of difference.

answer:Given that a < b < c and a = 0, we can deduce that b is the median of the three integers because it is the middle value. Let's denote the average of the three integers as A. According to the problem, the average is exactly 4 times the median (which is b). So we can write the equation: A = 4b The average of the three integers a, b, and c is the sum of the integers divided by 3: A = (a + b + c) / 3 Since a = 0, we can simplify this to: A = (0 + b + c) / 3 A = (b + c) / 3 Now we can equate the two expressions for A: (b + c) / 3 = 4b To solve for c, we multiply both sides by 3: b + c = 12b Subtract b from both sides to isolate c: c = 12b - b c = 11b Now we want to find the value of c / b: c / b = (11b) / b Since b is not equal to zero (because a < b < c and a = 0), we can simplify this to: c / b = 11 Therefore, the value of c / b is boxed{11} .

answer:To evaluate each option against the proposition "If angle 1 + angle 2 = 90^{circ}, then angle 1 neq angle 2", we proceed as follows: **Option A: angle 1 = angle 2 = 45^{circ}** - Condition Check: angle 1 + angle 2 = 45^{circ} + 45^{circ} = 90^{circ} - Conclusion Check: angle 1 = angle 2, which contradicts the conclusion angle 1 neq angle 2. Thus, option A satisfies the condition but does not satisfy the conclusion, making it a valid counterexample. **Option B: angle 1 = 50^{circ}, angle 2 = 50^{circ}** - Condition Check: angle 1 + angle 2 = 50^{circ} + 50^{circ} = 100^{circ} neq 90^{circ} - Conclusion Check: angle 1 = angle 2, which is irrelevant since the condition is not met. Option B does not satisfy the initial condition, making it not a valid counterexample. **Option C: angle 1 = 50^{circ}, angle 2 = 40^{circ}** - Condition Check: angle 1 + angle 2 = 50^{circ} + 40^{circ} = 90^{circ} - Conclusion Check: angle 1 neq angle 2, which satisfies both the condition and the conclusion. Option C satisfies both the condition and the conclusion, making it not a valid counterexample. **Option D: angle 1 = 40^{circ}, angle 2 = 40^{circ}** - Condition Check: angle 1 + angle 2 = 40^{circ} + 40^{circ} = 80^{circ} neq 90^{circ} - Conclusion Check: angle 1 = angle 2, which is irrelevant since the condition is not met. Option D does not satisfy the initial condition, making it not a valid counterexample. Therefore, the correct choice that serves as a counterexample to the proposition is: boxed{A}