answer:1. **Base Case:** For ( n = 1 ), the triangle ( triangle C_1B_1C_2 ) is formed by dividing ( AB ) into 1 part and ( AC ) into 2 parts. It is easy to see that the area of ( triangle C_1B_1C_2 ) is half of the area of ( triangle ABC ). 2. **Induction Hypothesis:** Assume that for ( n = k ), the shaded area is ( frac{1}{2} ) of the area of ( triangle ABC ). 3. **Induction Step:** For ( n = k+1 ), we need to show that the shaded area is still ( frac{1}{2} ) of the area of ( triangle ABC ). - The triangles ( triangle C_1B_1C_2, triangle C_2B_2C_3, ldots, triangle C_kB_kC_{k+1} ) are already shaded and their total area is ( frac{1}{2} ) of the area of ( triangle ABC ) by the induction hypothesis. - We need to add the area of the new triangle ( triangle C_{k+1}B_{k+1}C_{k+2} ). 4. **Area Calculation:** - The area of ( triangle C_{k+1}B_{k+1}C_{k+2} ) can be calculated as follows: [ text{Area of } triangle C_{k+1}B_{k+1}C_{k+2} = frac{1}{k+2} times text{Area of } triangle ABC ] - The total shaded area for ( n = k+1 ) is: [ left( frac{1}{2} times text{Area of } triangle ABC right) + left( frac{1}{k+2} times text{Area of } triangle ABC right) ] - Simplifying the expression: [ frac{1}{2} times text{Area of } triangle ABC + frac{1}{k+2} times text{Area of } triangle ABC = frac{k+2}{2(k+2)} times text{Area of } triangle ABC = frac{1}{2} times text{Area of } triangle ABC ] 5. **Conclusion:** By induction, the shaded area is always ( frac{1}{2} ) of the area of ( triangle ABC ) for any ( n ). (blacksquare) The final answer is ( boxed{ frac{1}{2} } ) of the area of (triangle ABC).

answer:Let's denote the cost of one pen as C and the number of pens sold as N. The trader gains the cost of 30 pens, which means his gain is 30C. The gain percentage is given as 33.33333333333333%, which can also be written as 1/3 or 33 1/3%. The gain percentage is calculated as (Gain / Total Cost) * 100. Since the gain is 30C, we can write the equation as: (30C / (N * C)) * 100 = 33.33333333333333 We can simplify this by canceling out C from the numerator and denominator: (30 / N) * 100 = 33.33333333333333 Now, we can solve for N: 30 * 100 = 33.33333333333333 * N 3000 = 33.33333333333333 * N To find N, we divide both sides by 33.33333333333333: N = 3000 / 33.33333333333333 N = 90 Therefore, the trader sold boxed{90} pens.

answer:To determine the coordinates of the point M(1,2) with respect to the x-axis, we reflect the point across the x-axis. This changes the sign of the y-coordinate while keeping the x-coordinate the same. Therefore, the transformation can be detailed as follows: - Original coordinates: (1,2) - Reflect across the x-axis: (1, -2) This means the coordinates of the point M(1,2) with respect to the x-axis are boxed{(1,-2)}.

answer:To find the domain of the function f(x) = sqrt{x-1} + frac{1}{x-2}, we need to consider the conditions under which the function is defined. 1. The square root function sqrt{x-1} requires that its argument x-1 must be greater than or equal to 0. This gives us the inequality: [x - 1 geq 0] Solving for x, we find: [x geq 1] 2. The fraction frac{1}{x-2} requires that its denominator x-2 must not be equal to 0 to avoid division by zero. This gives us the condition: [x - 2 neq 0] Solving for x, we find: [x neq 2] Combining these conditions, we find that the domain of f(x) is all x such that x geq 1 and x neq 2. Therefore, the domain of the function f(x) is: [boxed{{x | x geq 1 text{ and } x neq 2}}]