answer:Since (1+x)+(1+x)^{2}+(1+x)^{3}+...+(1+x)^{n}=a_{0}+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n}, Let x=1, we have 2+2^{2}+2^{3}+...+2^{n}=a_{0}+a_{1}+a_{2}+...+a_{n}=126, which implies frac {2(1-2^{n})}{1-2}=126, 2^{n+1}=128, thus n=6. According to the general term formula of (( sqrt {x}- frac {1}{ sqrt {x}})^{n})=( sqrt {x}- frac {1}{ sqrt {x}})^{6}, we have T_{r+1}= C_{ 6 }^{ r }⋅(-1)^{r}⋅x^{3-r}. Let 3-r=0, we find r=3, thus the constant term in the expansion is (- C_{ 6 }^{ 3 }=-20). Therefore, the answer is boxed{D}. By using the given conditions, we first determined n=6, then we utilized the general term formula of binomial expansion to find the constant term in the expansion of (( sqrt {x}- frac {1}{ sqrt {x}})^{n}). This problem mainly tests the application of the binomial theorem, binomial expansion's general term formula, finding the coefficient of a certain term, and the properties of binomial coefficients. It is a basic problem.

answer:We will prove the statement by induction on the number of students n in the class. 1. **Base Case:** For ( n = 3 ), the statement is obvious. 2. **Inductive Step:** Assume the statement holds for all classes with ( n leq N ) students. We will prove it holds for a class with ( n = N+1 ) students. 3. Suppose the statement is true if there is exactly one silent student. Let's assume there are at least two silent students. 4. Identify a silent student ( A ) and his friends – the chatterers ( B_1, B_2, ldots, B_k ). Consider the remaining ( n-1-k ) students. By the inductive hypothesis, there exists a group ( M ) within this remaining set where each chatterer in ( M ) has an odd number of silent friends and ( M ) includes at least ( frac{ leftlceil frac{n-1-k}{2} rightrceil }{2} ) students. 5. Let us consider the chatterers ( B_1, B_2, ldots, B_m ) who are friends with an odd number of silent students from ( M ), and ( B_{m+1}, B_{m+2}, ldots, B_k ) who are friends with an even number of silent students. - If ( m geq frac{k+1}{2} ): - Add the chatterers ( B_1, B_2, ldots, B_m ) to the group ( M ). - If ( m < frac{k+1}{2} ): - Add the chatterers ( B_{m+1}, B_{m+2}, ldots, B_k ) and the silent student ( A ) to the group ( M ). 6. In both cases, we form a group of students satisfying the conditions of the problem: each chatterer in the group has an odd number of silent friends in the group. Additionally, the group size includes at least ( frac{(n-1)}{2} ) students. Thus, by inductive reasoning, the teacher can invite at least half of the class such that all chatterers will be silent. (blacksquare)

answer:Adam spent 81 on the ferris wheel, and each ticket costs 9. To find out how many tickets he used for the ferris wheel, we divide the total amount spent by the cost per ticket: 81 / 9 per ticket = 9 tickets These 9 tickets were used for the ferris wheel. After riding the ferris wheel, Adam had 4 tickets left. To find out the total number of tickets he bought initially, we add the tickets used for the ferris wheel to the tickets he had left: 9 tickets (used for the ferris wheel) + 4 tickets (left) = 13 tickets So, Adam bought boxed{13} tickets initially.

answer:Let the roots of the quadratic be z = j + ki and overline{z} = j - ki, where j and k are real numbers. Using Vieta’s formulas, which relate the sums and products of roots to the coefficients of the polynomial: 1. z + overline{z} = (j + ki) + (j - ki) = 2j, and by the coefficient of z we have: [ -2j = 6 + pi ] Thus, for this to be real (since 2j is real), p must be zero. 2. z overline{z} = (j + ki)(j - ki) = j^2 + k^2, and matching with constant term: [ j^2 + k^2 = 13 + qi ] Again, for this term to be real, q must be zero. Combining both results, we conclude (p, q) = boxed{(0, 0)}.