answer:1. **Base Case:** For ( n = 1 ), the only permutation of ({1}) is the identity permutation, which can be written as the product of zero transpositions. Thus, ( c_{1,0} = 1 ) and ( c_{1,m} = 0 ) for ( m geq 1 ). Therefore, [ sum_{m=0}^{infty} c_{1,m} t^m = 1 = prod_{i=1}^{1} (1 + t + cdots + t^{i-1}) = 1. ] The base case holds. 2. **Inductive Hypothesis:** Assume that for some ( n in mathbb{N} ), [ sum_{m=0}^{infty} c_{n,m} t^m = prod_{i=1}^{n} (1 + t + cdots + t^{i-1}). ] 3. **Inductive Step:** We need to show that [ sum_{m=0}^{infty} c_{n+1,m} t^m = prod_{i=1}^{n+1} (1 + t + cdots + t^{i-1}). ] Consider the set ({1, ldots, n+1}). A permutation of this set can be written as the product of ( m ) transpositions of the form ((i, i+1)) for some ( i = 1, ldots, n). Let ( pi ) be a permutation of ({1, ldots, n+1}) that can be written as the product of ( m ) such transpositions. We can categorize these permutations based on the position of ( n+1 ): - If ( n+1 ) is in the ( (n+1) )-th position, the remaining ( n ) elements form a permutation of ({1, ldots, n}) that can be written as the product of ( m ) transpositions. - If ( n+1 ) is in the ( n )-th position, the remaining ( n ) elements form a permutation of ({1, ldots, n}) that can be written as the product of ( m-1 ) transpositions. - Continue this pattern until ( n+1 ) is in the first position, where the remaining ( n ) elements form a permutation of ({1, ldots, n}) that can be written as the product of ( m-n ) transpositions. Therefore, we have: [ c_{n+1,m} = c_{n,m} + c_{n,m-1} + cdots + c_{n,m-n}. ] 4. **Generating Function:** Consider the generating function: [ sum_{m=0}^{infty} c_{n+1,m} t^m = sum_{m=0}^{infty} left( c_{n,m} + c_{n,m-1} + cdots + c_{n,m-n} right) t^m. ] This can be rewritten as: [ sum_{m=0}^{infty} c_{n+1,m} t^m = left( sum_{m=0}^{infty} c_{n,m} t^m right) left( 1 + t + t^2 + cdots + t^n right). ] By the inductive hypothesis, we have: [ sum_{m=0}^{infty} c_{n,m} t^m = prod_{i=1}^{n} (1 + t + cdots + t^{i-1}). ] Therefore, [ sum_{m=0}^{infty} c_{n+1,m} t^m = left( prod_{i=1}^{n} (1 + t + cdots + t^{i-1}) right) (1 + t + cdots + t^n). ] Simplifying, we get: [ sum_{m=0}^{infty} c_{n+1,m} t^m = prod_{i=1}^{n+1} (1 + t + cdots + t^{i-1}). ] By induction, the statement holds for all ( n in mathbb{N} ). (blacksquare)

answer:Given f(x) = x(1 - x^2) = x - x^3, We find the first derivative f'(x) = 1 - 3x^2, By setting f'(x) = 0, we obtain x = frac{sqrt{3}}{3} or x = -frac{sqrt{3}}{3} (which is rejected as it is outside the given interval), Now, let's compute the function values at critical points and endpoints: f(0) = 0, f(frac{sqrt{3}}{3}) = frac{sqrt{3}}{3}(1 - frac{1}{3}) = frac{2sqrt{3}}{9}, f(1) = 0, Therefore, the maximum value of f(x) = x(1 - x^2) on the interval [0, 1] is boxed{frac{2sqrt{3}}{9}}.

answer:Given that x^{m}=5 and x^{n}=frac{1}{4}, we want to find the value of x^{2m-n}. First, we observe that x^{2m-n} can be rewritten using the properties of exponents. Specifically, we can express x^{2m-n} as (x^{m})^{2}div x^{n}. This is because multiplying exponents when the bases are the same is equivalent to adding the exponents, and dividing is equivalent to subtracting the exponents. Given x^{m}=5, we can square both sides to get (x^{m})^{2}=5^{2}. This simplifies to (x^{m})^{2}=25. Next, we know that x^{n}=frac{1}{4}. To divide by x^{n} is the same as multiplying by its reciprocal, so we have 25div x^{n} = 25 times 4 because the reciprocal of frac{1}{4} is 4. Putting it all together, we have: x^{2m-n} = (x^{m})^{2}div x^{n} = 25 div frac{1}{4} = 25 times 4 = 100. Therefore, the correct answer is boxed{D}.

answer:The sum of all angles around a point in the circle should equal 360°. We have angles labeled as 3x^circ, 7x^circ, 4x^circ, 2x^circ, and x^circ. Setting up the equation based on these: [ 3x + 7x + 4x + 2x + x = 360^circ ] [ 17x = 360^circ ] Dividing both sides by 17 to find x: [ x = frac{360}{17} ] Therefore, [ x = boxed{frac{360}{17}^circ} ] Conclusion: Given that the new angles also properly sum to 360°, the new problem statement fits the calculation requirement, and is consistent with the geometric principles involved.