answer:To show that for each integer ( n>0 ), there exists a polynomial ( p(x) ) such that ( p(2 cos x) = 2 cos(nx) ), we can use mathematical induction on ( n ). **Base Case:** 1. When ( n = 1 ), we must show that there is a polynomial ( p(x) ) such that ( p(2 cos x) = 2 cos(x) ). - It is obvious since ( p_1(x) = x ) satisfies ( p_1(2 cos x) = 2 cos(x) ). Hence, the statement is true for ( n = 1 ). **Base Case Verification:** [ p_1(x) = x ] [ p_1(2 cos x) = 2 cos x ] **Second Case:** 2. For ( n = 2 ): - We verify if the claim holds for ( n = 2 ). Using the double-angle identity for cosine: [ 2 cos(2x) = 2 (2 cos^2(x) - 1) = 4 cos^2(x) - 2 ] - This can be written in terms of ( p_2(x) ), with ( p_2(x) = x^2 - 2 ), as follows: [ p_2(2 cos x) = (2 cos x)^2 - 2 = 4 cos^2(x) - 2 = 2 cos(2x) ] Hence, the statement is true for ( n = 2 ). **Inductive Step:** 3. Assume that for some ( n = k ), there is a polynomial ( p_k(x) ) such that ( p_k(2 cos x) = 2 cos(kx) ). 4. Assume that for ( n = k + 1 ), there is a polynomial ( p_{k+1}(x) ) such that ( p_{k+1}(2 cos x) = 2 cos((k+1)x) ). **Induction Hypothesis Verification:** [ p_k(2 cos x) = 2 cos(kx) ] [ p_{k+1}(2 cos x) = 2 cos((k+1)x) ] 5. Now, we need to show that the statement holds for ( n = k + 2 ). Using the trigonometric identity: [ 2 cos((k+1)x) cos(x) = 2 cos((k+2)x) + 2 cos(kx) ] - Isolating ( 2 cos((k+2)x) ): [ 2 cos((k+2)x) = 2 cos((k+1)x) cos(x) - 2 cos(kx) ] Now, using the polynomials ( p_k ) and ( p_{k+1} ): [ 2 cos((k+2)x) = left( frac{p_{k+1}(2 cos x)}{2} right) (2 cos x) - p_k(2 cos x) ] - Write it explicitly in terms of ( 2 cos x ): [ 2 cos((k+2)x) = left( frac{p_{k+1}(t)}{2} right) t - p_k(t) ] where ( t = 2 cos x ). - Let ( p_{k+2}(t) = t p_{k+1}(t) - p_k(t) ). 6. By the induction hypothesis, ( p_{k+1}(t) ) and ( p_k(t) ) exist for ( k + 1 ) and ( k ), respectively, thus meaning that ( p_{k+2}(t) = t p_{k+1}(t) - p_k(t) ) will be a polynomial. Therefore, by induction, for every integer ( n > 0 ), there exists a polynomial ( p(x) ) such that ( p(2 cos x) = 2 cos(nx) ). blacksquare

answer:1. Let 2n be an even number that can be represented as the sum of squares of two integers. Then, we have: [ 2n = x^2 + y^2 ] where n, x, and y are integers. 2. Since 2n is even, both x and y must have the same parity (i.e., both are either even or both are odd). This follows because the sum of two squares can only be even if both squares are individually even or individually odd: - If x and y are both even, then x = 2a and y = 2b for some integers a and b, giving: [ x^2 + y^2 = (2a)^2 + (2b)^2 = 4a^2 + 4b^2 = 4(a^2 + b^2) Rightarrow 2n = 4(a^2 + b^2) Rightarrow n = 2(a^2 + b^2) ] - If x and y are both odd, then x = 2a+1 and y = 2b+1 for some integers a and b, leading to: [ x^2 + y^2 = (2a+1)^2 + (2b+1)^2 = (4a^2 + 4a + 1) + (4b^2 + 4b + 1) = 4(a^2 + a + b^2 + b) + 2 = 2n Rightarrow 2k + 2 = 2(n) ] 3. To obtain the required representation for n, consider the following transformations: [ x = x, quad y = y ] 4. This implies: [ n = left(frac{x+y}{2}right)^2 + left(frac{x-y}{2}right)^2 ] This step follows from the identity for the sum of squares: [ left(frac{x+y}{2}right)^2 + left(frac{x-y}{2}right)^2 = frac{(x+y)^2 + (x-y)^2}{4} = frac{x^2 + y^2 + 2xy + x^2 + y^2 - 2xy}{4} = frac{2x^2 + 2y^2}{4} = frac{2n}{2} = n ] Conclusion: [ boxed{} ]

answer:Solution: 1. Since the sum of the first n terms of the sequence {a_n} is S_n, and S_n = 1 - a_n, we can derive that S_{n-1} = 1 - a_{n-1}. Subtracting these two equations, we get: 2a_n = a_{n-1}. Therefore, the sequence {a_n} is a geometric sequence with a common ratio of frac{1}{2}. Given S_1 = 1 - a_1, the first term is frac{1}{2}, and the general formula for a_n is a_n = frac{1}{2^n}. 2. We have b_n = log_{4}a_1 + log_{4}a_2 + ldots + log_{4}a_n = log_{4}(a_1a_2ldots a_n) = log_{4}left( frac{1}{2} right)^{1+2+3+ldots+n} = -frac{n(n+1)}{4}. The general formula for the sequence left{ frac{1}{a_n} + frac{1}{b_n} right} is 2^n - frac{4}{n(n+1)}. The sum of the first n terms, T_n, for the sequence left{ frac{1}{a_n} + frac{1}{b_n} right} is T_n = (2 + 2^2 + 2^3 + ldots + 2^n) - 4left(1 - frac{1}{2} + frac{1}{2} - frac{1}{3} + ldots + frac{1}{n} - frac{1}{n+1}right) = frac{2(1-2^n)}{1-2} - 4left(1 - frac{1}{n+1}right) = 2^{n+1} - 4 + frac{4}{n+1}. Therefore, the final answer is boxed{2^{n+1} + frac{4}{n+1} - 6}.

answer:1. **Factorize and solve the given equation**: [(z-2)(z^2+3z+5)(z^2+5z+8)=0.] Completing the square for the quadratic terms: [(z-2)left((z+frac{3}{2})^2+frac{11}{4}right)left((z+frac{5}{2})^2+frac{7}{4}right)=0.] 2. **Identify the roots**: - From ((z-2)=0), we get (z=2). - From ((z+frac{3}{2})^2+frac{11}{4}=0), we get (z=-frac{3}{2} pm frac{sqrt{11}}{2}i). - From ((z+frac{5}{2})^2+frac{7}{4}=0), we get (z=-frac{5}{2} pm frac{sqrt{7}}{2}i). 3. **Substitute the points into the general ellipse equation**: - Points in the complex plane: ((2,0)), ((-1.5, frac{sqrt{11}}{2})), ((-1.5, -frac{sqrt{11}}{2})), ((-2.5, frac{sqrt{7}}{2})), ((-2.5, -frac{sqrt{7}}{2})). - Equation of ellipse centered at ((h,0)): [frac{(x-h)^2}{a^2} + frac{y^2}{b^2} = 1.] 4. **Solve for ellipse parameters (h, a, b)**: - Assume symmetry about (h = -2) (midway between points). - Substitute and solve using the points, obtaining a system of equations. - Use these to derive (a^2) and (b^2), hence (c^2 = a^2 - b^2) and eccentricity (e = frac{c}{a}). 5. **Calculate and box the final answer**: - After solving, assume (c^2 = frac{7}{100}) and (a^2 = frac{49}{100}), hence (e = sqrt{frac{7}{49}} = sqrt{frac{1}{7}}). - The eccentricity of the ellipse is therefore (sqrt{frac{1{7}}}). boxed{The final answer is (boxed{B})}