answer:# Solution: Part (1): Find the general formula for {a_{n}} Given that a_{1}=1 and the sequence left{frac{S_{n}}{a_{n}}right} forms an arithmetic sequence with a common difference of frac{1}{3}, we have: - The n-th term of this arithmetic sequence can be expressed as frac{S_{n}}{a_{n}} = 1 + frac{1}{3}(n-1) = frac{1}{3}n + frac{2}{3}. This leads to the equation for S_{n}: S_{n} = frac{1}{3}na_{n} + frac{2}{3}a_{n} quad text{(1)} For n geqslant 2, we can write a similar equation for S_{n-1}: S_{n-1} = frac{1}{3}(n-1)a_{n-1} + frac{2}{3}a_{n-1} quad text{(2)} Subtracting equation (2) from equation (1) gives us: a_{n} = frac{1}{3}na_{n} - frac{1}{3}(n-1)a_{n-1} Rearranging, we find: left(n-1right)a_{n} = left(n+1right)a_{n-1} This simplifies to a recursive relationship between consecutive terms: frac{a_{n}}{a_{n-1}} = frac{n+1}{n-1} By applying this relationship repeatedly, starting from a_{1}=1, we find: frac{a_{n}}{a_{1}} = frac{n(n+1)}{2} Therefore, the general formula for {a_{n}} is: boxed{a_{n} = frac{n(n+1)}{2}} Part (2): Prove that frac{1}{{a_1}}+frac{1}{{a_2}}+ldots +frac{1}{{a_n}} lt 2 Given the formula for a_{n}: a_{n} = frac{n(n+1)}{2} The reciprocal of a_{n} can be expressed as: frac{1}{a_{n}} = frac{2}{n(n+1)} = 2left(frac{1}{n} - frac{1}{n+1}right) Summing the reciprocals from 1 to n gives us a telescoping series: frac{1}{a_{1}} + frac{1}{a_{2}} + ldots + frac{1}{a_{n}} = 2left(1 - frac{1}{2} + frac{1}{2} - frac{1}{3} + ldots + frac{1}{n} - frac{1}{n+1}right) This simplifies to: 2left(1 - frac{1}{n+1}right) < 2 Therefore, we have proven that: boxed{frac{1}{{a_1}}+frac{1}{{a_2}}+ldots +frac{1}{{a_n}} lt 2}

answer:**Analysis** This question tests the method of solving inequalities, which is a basic question. Solve the inequality system begin{cases}(1−x)(x+1)leqslant 0 xneq -1end{cases}, from which the solution set of the inequality can be obtained. **Solution** Given begin{cases}(1−x)(x+1)leqslant 0 xneq -1end{cases}, therefore xgeqslant 1 or x < -1. Therefore, the correct choice is boxed{text{D}}.

answer:1. **Understanding the setup**: Each sphere has a radius of 2 and is tangent to three orthogonal planes, meaning the center of each sphere is at (2, 2, 2) in its respective corner with sign modifications per quadrant. 2. **Forming a cube by sphere centers**: These centers are vertices of a cube with side length 4 (twice the radius, as centers are 2 + 2 = 4 units apart in each dimension). 3. **Calculating the space diagonal of the cube**: The formula for the space diagonal d of a cube with side s is d = ssqrt{3}. Substituting s = 4, we get: [ d = 4sqrt{3} ] 4. **Determining the diameter of the enclosing sphere**: The diameter of the sphere is the space diagonal of the cube plus twice the radius of the small spheres: [ text{Diameter} = 4sqrt{3} + 4 = 4sqrt{3} + 4 ] 5. **Finding the radius of the enclosing sphere**: The radius of the enclosing sphere is half of its diameter: [ text{Radius} = frac{4sqrt{3} + 4}{2} = 2sqrt{3} + 2 ] Thus, the radius of the smallest sphere centered at the geometric center of the room that contains these four spheres is 2sqrt{3 + 2}. The final answer is boxed{mathrm{D}} 2sqrt{3} + 2

answer:**Analysis** This problem tests our understanding of the properties of hyperbolas and how to find the equations of their asymptotes, making it a fundamental question. **Step-by-step Solution** 1. From the given information, we know the eccentricity e=frac{c}{a}=frac{sqrt{5}}{2}. 2. Squaring both sides, we have frac{c^{2}}{a^{2}}=frac{5}{4}. 3. Recall that for hyperbolas, c^2=a^2+b^2. Therefore, frac{c^{2}}{a^{2}}=frac{a^{2}+b^{2}}{a^{2}}=1+left(frac{b}{a}right)^{2}. 4. Comparing the two expressions of frac{c^{2}}{a^{2}}, we have 1+left(frac{b}{a}right)^{2}=frac{5}{4}. 5. Solving for frac{b}{a}, we get frac{b}{a}=frac{1}{2}. 6. The equations of the asymptotes of a hyperbola frac{x^{2}}{a^{2}}-frac{y^{2}}{b^{2}}=1 are given by y=pm frac{b}{a}x. 7. Substituting frac{b}{a}=frac{1}{2} into the equation, we obtain the asymptotes' equations: boxed{y=pm frac{1}{2}x}. Hence, the correct answer is C.