answer:# Problem: Can three distinct numbers of the form 2^{n} + 1, where n is a natural number, be consecutive terms of a geometric progression? 1. **Assume the Existence:** Let's assume there exist distinct natural numbers (k, m,) and (n) such that (2^{k} + 1, 2^{m} + 1,) and (2^{n} + 1) are consecutive terms of a geometric progression. 2. **Properties of Geometric Progression:** If (2^{k} + 1, 2^{m} + 1,) and (2^{n} + 1) are consecutive terms of a geometric progression, then: [ (2^{m} + 1)^2 = (2^{k} + 1)(2^{n} + 1) ] 3. **Expand and Simplify:** Expand the left-hand and right-hand sides of the equation: [ (2^{m} + 1)^2 = 2^{2m} + 2 cdot 2^{m} cdot 1 + 1^2 = 2^{2m} + 2^{m+1} + 1 ] and [ (2^{k} + 1)(2^{n} + 1) = 2^{k} cdot 2^{n} + 2^{k} cdot 1 + 2^{n} cdot 1 + 1 cdot 1 = 2^{k+n} + 2^{k} + 2^{n} + 1 ] 4. **Equate and Analyze:** Setting the expanded forms equal to each other: [ 2^{2m} + 2^{m+1} + 1 = 2^{k+n} + 2^{k} + 2^{n} + 1 ] 5. **Cancellation of Common Terms:** Subtract 1 from both sides: [ 2^{2m} + 2^{m+1} = 2^{k+n} + 2^{k} + 2^{n} ] 6. **Representation as Sums of Powers of 2:** According to the uniqueness of the representation of numbers as sums of different powers of 2 (based on the uniqueness of binary representation), the equation above can only hold if each corresponding power term on both sides matches. - The left side (2^{2m} + 2^{m+1}) has only two distinct power terms. - The right side (2^{k+n} + 2^{k} + 2^{n}) has three distinct power terms. This clear mismatch indicates that there cannot be an equality due to the different number of distinct powers of 2. 7. **Conclusion:** Therefore, it is impossible for three numbers of the form (2^{n} + 1), where (n) is a natural number, to be consecutive terms of a geometric progression. [ boxed{text{Cannot}} ]

answer:First, rewrite (frac{5}{4x^{-4}}) as ( frac{5x^4}{4} ) using the property ( x^{-a} = frac{1}{x^a} ). The expression becomes: [ frac{5}{4x^{-4}} cdot frac{2x^{3}}{3} = frac{5x^4}{4} cdot frac{2x^3}{3} ] Next, simplify the expression by multiplying the numerators and denominators: [ = frac{(5 cdot 2)(x^4 cdot x^3)}{4 cdot 3} = frac{10x^{4+3}}{12} ] Simplify the fraction: [ = frac{10x^7}{12} = frac{5x^7}{6} ] Thus, the simplified form is [ boxed{frac{5x^7}{6}} ]

answer:1. **Given Functional Equations:** [ f(2 + x) = f(2 - x) quad text{and} quad f(7 + x) = f(7 - x) ] These equations imply that ( f ) is symmetric about ( x = 2 ) and ( x = 7 ). 2. **Transforming the Equations:** Let's denote: [ a(x) = 4 - x quad text{and} quad b(x) = 14 - x ] We can rewrite the given functional equations as: [ f(x) = f(4 - x) quad text{and} quad f(x) = f(14 - x) ] 3. **Lemma:** We need to show that ( f(y) = 0 ) if and only if there exists a sequence of transformations using ( a ) and ( b ) that maps 0 to ( y ). 4. **Sequence of Transformations:** - Applying ( a ) twice returns the original value: ( a(a(x)) = x ). - Applying ( b ) twice returns the original value: ( b(b(x)) = x ). 5. **Generating Roots:** - Starting with ( x = 0 ): - Applying ( a ) and then ( b ): [ b(a(0)) = b(4) = 14 - 4 = 10 ] - Repeating this process: [ b(a(10)) = b(4 - 10) = b(-6) = 14 - (-6) = 20 ] - Generalizing, we get roots of the form ( 10j ) where ( j in mathbb{Z} ). - Applying ( b ) and then ( a ): [ a(b(0)) = a(14) = 4 - 14 = -10 ] - Repeating this process: [ a(b(-10)) = a(14 - (-10)) = a(24) = 4 - 24 = -20 ] - Generalizing, we get roots of the form ( -10j ) where ( j in mathbb{Z} ). 6. **Combining Results:** - Roots of the form ( 10j ) and ( 4 + 10j ) where ( j in mathbb{Z} ). 7. **Counting Roots in the Interval ([-1000, 1000]):** - For ( 10j ): [ -1000 leq 10j leq 1000 implies -100 leq j leq 100 ] This gives ( 201 ) roots. - For ( 4 + 10j ): [ -1000 leq 4 + 10j leq 1000 implies -1004 leq 10j leq 996 implies -100.4 leq j leq 99.6 ] Since ( j ) must be an integer, we have ( -100 leq j leq 99 ), giving ( 200 ) roots. 8. **Total Number of Roots:** [ 201 + 200 = 401 ] The final answer is (boxed{401}).

answer:**Analysis** This question examines the general formula of an arithmetic sequence. Finding the first term and the common difference is key to solving the problem, making it a basic question. Given the conditions, it's easy to find the sequence's first term and common difference, from which the general formula can be derived. **Solution** Let the common difference of the arithmetic sequence {a_n} be d, since a_n+a_{n+2}=4n+6, (1) then a_{n+2}+a_{n+4}=4(n+2)+6, (2) Subtracting (1) from (2) gives a_{n+4}-a_{n}=8, which means 4d=8, solving this gives d=2, Substituting n=1 into a_n+a_{n+2}=4n+6 gives 2a_1+4=10, solving this gives a_1=3, Therefore, the general formula is a_n=3+2(n-1)=2n+1, Hence, the answer is boxed{2n+1}.