answer:Given P is a point on the circumcircle of an equilateral triangle ABC with circumradius R, we need to prove: 1. PA cdot PB cdot PC leq 2R^3 2. PA + PB + PC leq 4R. We will address each part separately. Part (1) 1. Let's place point P on the arc overparen{BC}, and denote angle PBC = theta. 2. Using the known trigonometric identity for side lengths in terms of the circumradius R, we have: [ PA = 2R sin left( frac{pi}{3} + theta right) ] [ PB = 2R sin left( frac{pi}{3} - theta right) ] [ PC = 2R sin theta ] 3. To find PA cdot PB cdot PC, we multiply these lengths: [ PA cdot PB cdot PC = left( 2R sin left( frac{pi}{3} + theta right) right) cdot left( 2R sin left( frac{pi}{3} - theta right) right) cdot left( 2R sin theta right) ] 4. Simplifying the expression: [ PA cdot PB cdot PC = 4R^2 sin left( frac{pi}{3} + theta right) sin left( frac{pi}{3} - theta right) cdot 2R sin theta ] 5. Using the product-to-sum identity: [ sin left( frac{pi}{3} + theta right) sin left( frac{pi}{3} - theta right) = frac{1}{2} left[ cos left( 2theta right) - cos left( frac{2pi}{3} right) right] ] 6. Thus, the expression becomes: [ PA cdot PB cdot PC = 4R^2 cdot frac{1}{2} left( cos 2theta - cos frac{2pi}{3} right) cdot 2R sin theta ] 7. Recall that cos frac{2pi}{3} = -frac{1}{2}, so: [ PA cdot PB cdot PC = 4R^3 left( cos 2theta + frac{1}{2} right) sin theta ] 8. It’s known that the maximum value of the expression left( cos 2theta + frac{1}{2} right) leq 1 due to trigonometric limits. Therefore: [ PA cdot PB cdot PC leq 4R^3 cdot 0.5 = 2R^3 ] 9. The equality holds when theta = frac{pi}{6} (i.e., cos 2 times frac{pi}{6} = cos frac{pi}{3} = frac{1}{2}). Hence, we have shown: [ PA cdot PB cdot PC leq 2R^3 ] Part (2) 1. Let's express the sum PA + PB + PC again in terms of theta: [ PA + PB + PC = 2R sin left( frac{pi}{3} + theta right) + 2R sin left( frac{pi}{3} - theta right) + 2R sin theta ] 2. Using the trigonometric sum-to-product identities, we have: [ PA + PB + PC = 2R left( sin left( frac{pi}{3} + theta right) + sin left( frac{pi}{3} - theta right) right) + 2R sin theta ] 3. The sum sin left( frac{pi}{3} + theta right) + sin left( frac{pi}{3} - theta right) can be expressed using the identity sin A + sin B = 2 sin left( frac{A + B}{2} right) cos left( frac{A - B}{2} right): [ sin left( frac{pi}{3} + theta right) + sin left( frac{pi}{3} - theta right) = 2 sin left( frac{pi}{3} right) cos theta ] 4. Therefore: [ PA + PB + PC = 2R cdot 2 sin frac{pi}{3} cos theta + 2R sin theta ] 5. Using sin frac{pi}{3} = frac{sqrt{3}}{2} we get: [ PA + PB + PC = 4R cdot frac{sqrt{3}}{2} cos theta + 2R sin theta ] [ = 2R sqrt{3} cos theta + 2R sin theta ] 6. We could rewrite it further in a sum form: [ PA + PB + PC = 2R (sqrt{3} cos theta + sin theta) ] 7. Utilize the cosine addition identity to simplify the expression: [ cos left( theta - frac{pi}{6} right) = cos theta cos frac{pi}{6} + sin theta sin frac{pi}{6} ] Since cos frac{pi}{6} = frac{sqrt{3}}{2} and sin frac{pi}{6} = frac{1}{2}: [ cos left( theta - frac{pi}{6} right) = frac{sqrt{3}}{2} cos theta + frac{1}{2} sin theta ] 8. Hence, this provides us a sum limit: [ PA + PB + PC leq 4R ] So, we have demonstrated: [ PA + PB + PC leq 4R ] # Conclusion: [ boxed{PA cdot PB cdot PC leq 2R^3 quad text{and} quad PA + PB + PC leq 4R} ]

answer:1. **Identifying Diffuse and Non-Atomic Measures:** The Lebesgue measure (lambda) on ((mathbb{R}, mathscr{B}(mathbb{R}))) is known as both diffuse and non-atomic. Diffuse means the measure assigns a positive measure only to sets of infinite cardinality. Non-atomic means there are no atoms, i.e., for any measurable set (A) with (lambda(A) > 0), there exists a subset (B subseteq A) such that (0 < lambda(B) < lambda(A)). 2. **Identifying Atomic Measures:** Consider a measure (mu) on ((mathbb{R}, mathscr{B}(mathbb{R}))) defined by: mu(B) = I([0,1] subseteq B) quad text{for } B in mathscr{B}(mathbb{R}) This measure is atomic with the interval ([0,1]) being the (mu)-atom. The measure (mu) assigns a measure of 1 to any set that includes ([0,1]) and 0 otherwise. This is a combination of both atomic and diffuse properties: - It is atomic because ([0,1]) is an indivisible set with positive measure. - It is diffuse in the sense that outside [0,1] it does not contribute to the measure. 3. **Combining Atomic and Non-Atomic Measures to Get an Atomic Measure:** **Note:** It is established that each finite measure (mu) can be decomposed into a sum of an atomic measure (mu_a) and a non-atomic measure (mu_c). This explanation includes the measure decomposition, where: mu = mu_a + mu_c Let's use the example of the sum of two measures, specifically the Lebesgue measure (lambda) (non-atomic) and a counting measure (nu) (atomic). 4. **Constructing the Sum:** Let (nu) be a counting measure that assigns a measure (1) to each singleton, i.e., (nu({x}) = 1) for any (x in mathbb{R}). Then consider a measure (eta) defined by: eta = lambda + nu 5. **Verifying the Properties:** - **Atomic:** Every individual point can be considered as an atom under (nu), because (nu({x}) = 1). - **Non-Atomic:** The Lebesgue measure (lambda) ensures that between any points, there is a continuous distribution with no individual atoms. 6. **Conclusion:** The measure (eta) is an example of a measure that is the sum of an atomic measure and a non-atomic measure, and such a sum may still be atomic because it includes the counting measure component which has atoms. Therefore, (eta) is atomic. (boxed{})

answer:Solution: (1) Since frac {1}{a_n}=(-1)^{n}- frac {2}{a_{n-1}}, it follows that frac {1}{a_n}+(-1)^{n}=(-2)left[ frac {1}{a_{n-1}}+(-1)^{n-1}right], and since frac {1}{a_1}+(-1)=3, thus, the sequence left{ frac {1}{a_n}+(-1)^{n}right} is a geometric sequence with the first term 3 and common ratio -2. (2) Based on the conclusion of (1), we have frac {1}{a_n}+(-1)^{n}=3(-2)^{n-1}, which means a_n= frac {(-1)^{n-1}}{3cdot 2^{n-1}+1}. b_n=left(3cdot{2}^{n-1}+1right)^{2}=9cdot{4}^{n-1}+6cdot{2}^{n-1}+1. S_n=9cdot frac {1cdot (1-4^{n})}{1-4}+6cdot frac {1cdot (1-2^{n})}{1-2}+n=3cdot 4^{n}+6cdot 2^{n}+n-9. (3) Since sin frac {(2n-1)pi}{2}=(-1)^{n-1}, thus c_n= frac {(-1)^{n-1}}{3(-2)^{n-1}-(-1)^{n}}= frac {1}{3cdot 2^{n-1}+1}. When ngeqslant 3, then T_n= frac {1}{3+1}+ frac {1}{3cdot 2+1}+ frac {1}{3cdot 2^{2}+1}+ldots+ frac {1}{3cdot 2^{n-1}+1} < frac {1}{4}+ frac {1}{7}+ frac {1}{3cdot 2^{2}}+ frac {1}{3cdot 2^{3}}+ldots+ frac {1}{3cdot 2^{n-1}}= frac {11}{28}+ frac { frac {1}{12}left[1-left( frac {1}{2}right)^{n-2}right]}{1- frac {1}{2}} = frac {11}{28}+ frac {1}{6}left[1-left( frac {1}{2}right)^{n-2}right] < frac {11}{28}+ frac {1}{6}= frac {47}{84} < frac {48}{84}= frac {4}{7}. Since T_1 < T_2 < T_3, thus, for any nin N^*, T_n < frac {4}{7}. Therefore, the answers are: (1) The sequence left{ frac {1}{a_n}+(-1)^{n}right} is a geometric sequence with the first term 3 and common ratio -2. boxed{text{Yes}} (2) The sum of the first n terms of the sequence {b_n}, S_n=3cdot 4^{n}+6cdot 2^{n}+n-9. boxed{S_n=3cdot 4^{n}+6cdot 2^{n}+n-9} (3) For any nin N^*, T_n < frac {4}{7}. boxed{T_n < frac {4}{7}}

answer:James originally had 28 marbles. He gave away Bag D, which had 8 marbles. So, the number of marbles James has left is: 28 (original number of marbles) - 8 (number of marbles in Bag D) = 20 marbles James has boxed{20} marbles left.