answer:Define A_1(n), A_2(n), and A_3(n) similar to the previous problem: - A_1(n): number of valid strings of length n ending in 1 but not in 111. - A_2(n): number of valid strings of length n ending in a single 0. - A_3(n): number of valid strings of length n ending in double zero. Initial conditions: - A_1(1) = 1, A_2(1) = 1, A_3(1) = 0. - A_1(2) = 2, A_2(2) = 1, A_3(2) = 1. - A_1(3) = 4 (consider 001, 101, 010, 110), A_2(3) = 1, A_3(3) = 1. Recursive relations: - A_1(n) = A_1(n - 1) + A_2(n - 1) + A_3(n - 1) - A_1(n - 3) (excluding strings that result in ending with 111). - A_2(n) = A_1(n - 1). - A_3(n) = A_2(n - 1). Total number of valid sequences: mathcal{T}_n = A_1(n) + A_2(n) + A_3(n) Calculate mathcal{T}_9: - mathcal{T}_4 = A_1(4) + A_2(4) + A_3(4) = 7 + 4 + 1 = 12 - mathcal{T}_5 = 12 + 7 + 4 = 23 - mathcal{T}_6 = 23 + 12 + 7 = 42 - mathcal{T}_7 = 42 + 23 + 12 = 77 - mathcal{T}_8 = 77 + 42 + 23 = 142 - mathcal{T}_9 = 142 + 77 + 42 - 4 (to adjust for the strings ending in 111) = 257 Thus, boxed{257} strings satisfy the conditions for mathcal{T}_9.

answer:1. Given the parabola ( y^2 = 4x ) with its focus at ( F ). 2. Consider a line passing through the focus ( F ) with a slope ( theta ) where ( 0 < theta < frac{pi}{2} ). 3. The equation of the line with slope ( k = tan(theta) ) passing through focus ( F(1, 0) ) is: [ y = k(x - 1) ] 4. The points ( A ) and ( B ) are the intersections of this line and the parabola. Substituting ( y = k(x - 1) ) into the parabola's equation ( y^2 = 4x ), we get: [ (k(x - 1))^2 = 4x ] [ k^2(x - 1)^2 = 4x ] [ k^2 (x^2 - 2x + 1) = 4x ] [ k^2 x^2 - 2k^2 x + k^2 = 4x ] Rearranging terms: [ k^2 x^2 - (2k^2 + 4) x + k^2 = 0 ] 5. Solving this quadratic equation for ( x ): [ x = frac{(2k^2 + 4) pm sqrt{(2k^2 + 4)^2 - 4 k^2 cdot k^2}}{2k^2} ] [ x = frac{2k^2 + 4 pm sqrt{4k^4 + 16 k^2}}{2k^2} ] [ x = frac{2k^2 + 4 pm 2ksqrt{k^2 + 4}}{2k^2} ] Simplifying further: [ x = frac{k^2 + 2 pm ksqrt{k^2 + 4}}{k^2} ] Therefore, [ x_1 + x_2 = frac{k^2 + 2 + k sqrt{k^2 + 4}}{k^2} + frac{k^2 + 2 - k sqrt{k^2 + 4}}{k^2} = frac{2(k^2 + 2)}{k^2} = 2 + frac{4}{k^2} ] 6. The sum of ( y )-coordinates ( y_1 + y_2 = frac{4}{k} ) and product ( y_1 y_2 = -4 ). 7. The line ( AO ) through point ( A left(x_1, y_1right)) and the origin ( O(0,0)) has slope ( y_1 / x_1 ); thus, its equation is: [ y = frac{y_1}{x_1} x ] Given the parabola equation ( y^2 = 4x ), substituting: [ y = frac{4}{y_1} x ] Hence, the intersection with the directrix ( x = -1 ) gives: [ B' = left(-1, -frac{4}{y_1}right) ] 8. Similarly, the intersection of ( BO ) with the directrix at ( A' ): [ A' = left(-1, -y_2right) ] Thus, coordinates: [ B = (x_2, y_2), B' = (-1, -frac{4}{y_1}), A = (x_1, y_1), A' = (-1, -y_2) ] 9. Quadrilateral ( ABB'A' ) is a trapezoid. 10. Area of trapezoid: [ S_{ABB'A'} = frac{1}{2} left( |AA'| + |BB'| right) |A'B'| ] 11. (AA' = |A-A'| = |y_1 + frac{4}{y_1}|), [ BB' = |B-B'| = |y_2 + y_2| = 2 cdot left| frac{4}{y_1} right| ] (AB = |A-A| = (y_1 + y_2)^2), Concluding the area as: [ 8(1 + frac{1}{k^2})^{frac{3}{2}} = 8 sec^3(theta) ] 12. Given secant identity: [ = frac{8}{sin^3 (theta)} ] Conclusion: [ boxed{frac{8}{sin^3 (theta)}} ]

answer:(1) Let the common difference of the arithmetic sequence {a_n} be d. From the given information, we have a_2 + a_5 + a_8 = 33, which implies a_5 = 11. Also, (11 - 4d + 2)(11 - 2d + 13) = (11 - 3d + 5)^2, solving this equation yields d = 2 or d = -28 (discard the latter), a_1 = a_5 - 4d = 3, thus a_n = a_1 + (n-1)d = 2n + 1. Furthermore, b_1 = a_1 + 2 = 5, b_2 = a_2 + 5 = 10, thus q = 2, therefore b_n = 5 times 2^{n-1}. (2) c_n = frac{a_n}{b_n} + 1 = frac{2n + 1}{5 times 2^{n-1}} + 1, thus T_n = frac{3}{5 cdot 2^0} + frac{5}{5 cdot 2} + frac{7}{5 cdot 2^2} + ldots + frac{2n + 1}{5 cdot 2^{n-1}} + n, frac{1}{2}T_n = frac{3}{5 cdot 2} + frac{5}{5 cdot 2^2} + ldots + frac{2n + 1}{5 cdot 2^n} + frac{1}{2}n, Subtracting the two equations yields frac{1}{2}T_n = frac{1}{5}left[ frac{3}{2^0} + frac{2}{2} + frac{2}{2^2} + ldots + frac{2}{2^{n-1}}right] - frac{2n + 1}{5 cdot 2^n} + frac{1}{2}n, thus T_n = 2 + n - frac{2n + 5}{5 cdot 2^{n-1}}. Therefore, the final answers are: (1) The general formula for {a_n} is boxed{a_n = 2n + 1} and for {b_n} is boxed{b_n = 5 times 2^{n-1}}. (2) The sum of the first n terms of the sequence {c_n}, T_n, is boxed{T_n = 2 + n - frac{2n + 5}{5 cdot 2^{n-1}}}.

answer:1. Divide the square park with side length 2 into sixteen smaller squares, each with side length frac{1}{2}. 2. There are nine potted plants, and they are distributed across these sixteen smaller squares. 3. By the Pigeonhole Principle, at least one of these squares must contain at least two plants (since placing one plant per square would use up only up to eight squares). 4. The maximum distance between two points within one of these smaller squares (which are frac{1}{2} by frac{1}{2}) would be the length of the diagonal. 5. The diagonal's length can be calculated using the Pythagorean theorem: sqrt{left(frac{1}{2}right)^2 + left(frac{1}{2}right)^2} = sqrt{frac{1}{4} + frac{1}{4}} = sqrt{frac{1}{2}} = frac{sqrt{2}}{2}. The smallest possible distance guaranteed between at least one pair of plants, given this placement, is boxed{frac{sqrt{2}}{2}}.