answer:1. **Base Case: ( n = 1 )** For ( n = 1 ), we have the function ( f_1(x) ). Given the conditions: [ f_1(x + y) = f_1(x) + y ] and [ f_1(-x) = -f_1(x) ] Let's analyze these conditions: - From ( f_1(x + y) = f_1(x) + y ), set ( y = 0 ): [ f_1(x) = f_1(x) + 0 implies f_1(0) = 0 ] - From ( f_1(-x) = -f_1(x) ), set ( x = 0 ): [ f_1(0) = -f_1(0) implies f_1(0) = 0 ] - Now, set ( y = -x ) in the first condition: [ f_1(x - x) = f_1(x) - x implies f_1(0) = f_1(x) - x implies 0 = f_1(x) - x implies f_1(x) = x ] Therefore, ( f_1(x) = x ). The base case is true. 2. **Inductive Step: Assume ( f_k(x_1, x_2, ldots, x_k) = frac{x_1 + x_2 + cdots + x_k}{k} ) is true for some ( k )** We need to show that ( f_{k+1}(x_1, x_2, ldots, x_{k+1}) = frac{x_1 + x_2 + cdots + x_{k+1}}{k+1} ). Given the conditions: [ f_{k+1}(f_k(x_1, ldots, x_k), ldots, f_k(x_1, ldots, x_k), x_{k+1}) = f_{k+1}(x_1, ldots, x_{k+1}) ] By the inductive hypothesis: [ f_k(x_1, x_2, ldots, x_k) = frac{x_1 + x_2 + cdots + x_k}{k} ] Substitute this into the condition: [ f_{k+1}left(frac{x_1 + x_2 + cdots + x_k}{k}, ldots, frac{x_1 + x_2 + cdots + x_k}{k}, x_{k+1}right) = f_{k+1}(x_1, x_2, ldots, x_{k+1}) ] Since ( f_{k+1} ) is symmetric, we can assume: [ f_{k+1}left(frac{x_1 + x_2 + cdots + x_k}{k}, ldots, frac{x_1 + x_2 + cdots + x_k}{k}, x_{k+1}right) = f_{k+1}(x_1, x_2, ldots, x_{k+1}) ] Now, consider the function ( f_{k+1} ) and the condition: [ f_{k+1}(x_1 + y, x_2 + y, ldots, x_{k+1} + y) = f_{k+1}(x_1, x_2, ldots, x_{k+1}) + y ] Let ( y = -frac{x_1 + x_2 + cdots + x_{k+1}}{k+1} ): [ f_{k+1}left(x_1 - frac{x_1 + x_2 + cdots + x_{k+1}}{k+1}, ldots, x_{k+1} - frac{x_1 + x_2 + cdots + x_{k+1}}{k+1}right) = f_{k+1}(x_1, x_2, ldots, x_{k+1}) - frac{x_1 + x_2 + cdots + x_{k+1}}{k+1} ] Since ( f_{k+1} ) is symmetric and the left-hand side becomes zero: [ f_{k+1}(0, 0, ldots, 0) = 0 ] Therefore: [ f_{k+1}(x_1, x_2, ldots, x_{k+1}) = frac{x_1 + x_2 + cdots + x_{k+1}}{k+1} ] By induction, ( f_n(x_1, x_2, ldots, x_n) = frac{x_1 + x_2 + cdots + x_n}{n} ) for all ( n ). (blacksquare) The final answer is ( boxed{ f_n(x_1, ldots, x_n) = frac{x_1 + x_2 + cdots + x_n}{n} } )

answer:To solve the problem, we start with the given condition in triangle ABC, which states 2acdot cos B = c. This equation is our starting point to deduce the nature of the triangle. 1. We use the Law of Sines to transform the initial condition: 2cos B = frac{c}{a} 2. According to the Law of Sines, we have: frac{sin A}{a} = frac{sin B}{b} = frac{sin C}{c} So, we can write: 2cos Bsin A = frac{c}{a}sin A = sin C 3. We can express sin C as a sum of angles: sin C = sin(A + B) 4. Expanding sin(A + B) using the sum of angles identity, we get: sin Acos B + cos Asin B = 2cos Bsin A 5. Rearranging the equation to bring similar terms to one side, we have: sin Acos B - cos Asin B = 0 6. This equation can be simplified using the sine of a difference formula: sin(A - B) = 0 7. The equation sin(A - B) = 0 implies that A - B = 0 or A = B. 8. Having A = B indicates that triangle ABC has at least two equal angles, making it an isosceles triangle. Therefore, the correct choice is boxed{B}: isosceles triangle.

answer:1. **Observe the Structure of the Numbers**: - First, we examine the structure of the numbers given: (underbrace{999999999 ldots 999}_{20089 text{ digits}})^{2007} and (underbrace{333333333 ldots 333}_{20083 text{ digits}})^{2007} The first number consists of 20089 digits, all of which are '9'. The second number consists of 20083 digits, all of which are '3'. 2. **Sum of Digits**: - For divisibility by 11, we use the rule that states: A number is divisible by 11 if the alternating sum of its digits is divisible by 11. 3. **Check Divisibility of (999...999)**: - The alternating sum of digits for the number with 20089 digits of '9': - We notice that '9' - '9' + '9' - '9' + ... = 9 (since the number of digits is odd). - Hence, the sum is effectively 9. 4. **Divisibility for (333...333)**: - The alternating sum of digits for the number with 20083 digits of '3': - Similarly, '3' - '3' + '3' - '3' + ... = 3 (since the number of digits is odd). - Hence, the sum is effectively 3. 5. **Powers of the Numbers**: - When raised to the power of 2007, both sums will be spread across all digits but since we are concerned with the property modulo 11, the multiplicity does not affect the alternating sum's property of the base number. - Therefore: [ (999...999)^{2007} equiv 0 pmod{11} quad text{because it is divisible by 11} ] [ (333...333)^{2007} equiv 0 pmod{11} quad text{because it likewise is divisible by 11} ] 6. **Difference**: - The difference of two numbers each congruent to 0 modulo 11 will also be divisible by 11: [ (999...999)^{2007} - (333...333)^{2007} equiv 0 - 0 equiv 0 pmod{11} ] # Conclusion: The remainder when the given expression is divided by 11 is: [ boxed{0} ]

answer:To simplify and then evaluate the given expression left(2x+yright)^{2}+left(x+yright)left(x-yright)-x^{2} with x=-1 and y=2, we follow these steps: 1. **Expand and simplify the expression:** First, we expand left(2x+yright)^{2}: [ left(2x+yright)^{2} = 4x^{2} + 4xy + y^{2} ] Next, we expand left(x+yright)left(x-yright): [ left(x+yright)left(x-yright) = x^{2} - y^{2} ] Substituting these expansions into the original expression gives: [ 4x^{2} + 4xy + y^{2} + x^{2} - y^{2} - x^{2} ] Combining like terms: [ = 4x^{2} + 4xy + y^{2} + x^{2} - y^{2} - x^{2} = 4x^{2} + 4xy ] 2. **Substitute x=-1 and y=2 into the simplified expression:** Substituting x=-1 and y=2 into 4x^{2} + 4xy: [ 4times (-1)^{2} + 4times (-1)times 2 = 4 - 8 ] Simplifying this gives: [ 4 - 8 = -4 ] Therefore, the simplified and evaluated expression is boxed{-4}.