answer:1. We begin with the series: [ left( frac{1}{2} + frac{1}{3} + cdots + frac{1}{2016} right) + left( frac{2}{3} + frac{2}{4} + cdots + frac{2}{2016} right) + left( frac{3}{4} + frac{3}{5} + cdots + frac{3}{2016} right) + cdots + left( frac{2014}{2015} + frac{2014}{2016} right) + frac{2015}{2016}. ] 2. Group the terms based on common denominators: [ frac{1}{2} + left( frac{1}{3} + frac{2}{3} right) + left( frac{1}{4} + frac{2}{4} + frac{3}{4} right) + cdots + left( frac{1}{2015} + cdots + frac{2014}{2015} right) + left( frac{1}{2016} + cdots + frac{2015}{2016} right). ] 3. Simplify each group by summing the fractions: [ frac{1}{2} + 1 + frac{3}{2} + cdots + frac{1 + 2 + cdots + 2014}{2015} + frac{1 + 2 + cdots + 2015}{2016}. ] Notice that the sums (1 + 2 + cdots + n) can be calculated using the formula for the sum of the first (n) natural numbers, ( S = frac{n(n+1)}{2} ). 4. Applying this formula, we get: [ frac{1}{2} + 1 + frac{3}{2} + cdots + frac{frac{1 cdot 2015}{2}}{2015} + frac{frac{1 cdot 2016}{2}}{2016}. ] 5. Simplify each term: [ frac{1}{2} + frac{2}{2} + frac{3}{2} + cdots + frac{2014}{2} + frac{2015}{2}. ] 6. This series amounts to: [ frac{1 + 2 + 3 + cdots + 2015}{2}. ] 7. Compute the sum of the sequence (1, 2, 3, ldots, 2015): [ S = sum_{k=1}^{2015} k = frac{2015 times 2016}{2}. ] 8. Include the factor ( frac{1}{2} ): [ frac{1}{2} times frac{2015 times 2016}{2} = frac{2015 times 2016}{4}. ] 9. Compute the final multiplication: [ 2015 times 2016 = 4062240, ] and [ frac{4062240}{4} = 1015560. ] Conclusion: [ boxed{1015560} ]

answer:To determine the maximum possible value of the surface area of a rectangular parallelepiped with a given diagonal length of 3, we follow these steps: 1. **Identify the given information and establish the relation between edges and diagonal**: Let the edges of the rectangular parallelepiped be a, b, and c. We know the length of the diagonal is 3, so: [ sqrt{a^2 + b^2 + c^2} = 3 ] Squaring both sides, we get: [ a^2 + b^2 + c^2 = 9 ] 2. **Define the formula for the surface area**: The surface area ( S ) of a rectangular parallelepiped is given by: [ S = 2(ab + bc + ca) ] 3. **Prove the inequality involving the product terms**: We need to prove that: [ ab + bc + ca leq a^2 + b^2 + c^2 ] Start by considering the following: [ (a - b)^2 + (b - c)^2 + (c - a)^2 geq 0 ] Expanding the squares, we have: [ a^2 - 2ab + b^2 + b^2 - 2bc + c^2 + c^2 - 2ca + a^2 geq 0 ] Simplifying this, we obtain: [ 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca geq 0 ] Dividing both sides by 2, we get: [ a^2 + b^2 + c^2 geq ab + bc + ca ] 4. **Calculate the maximum possible value**: Given that: [ a^2 + b^2 + c^2 = 9 ] From the inequality we just proved, it follows that: [ ab + bc + ca leq 9 ] Therefore, the maximum value of ( S ) is: [ S = 2(ab + bc + ca) leq 2 times 9 = 18 ] 5. **Determine the conditions for equality**: The equality ( ab + bc + ca = a^2 + b^2 + c^2 ) holds if and only if: [ a = b = c ] Given ( a^2 + b^2 + c^2 = 9 ) and ( a = b = c ), we have: [ 3a^2 = 9 implies a^2 = 3 implies a = sqrt{3} ] Therefore, the shape must be a cube with edge length ( sqrt{3} ). # Conclusion: The maximum possible value of the surface area of a rectangular parallelepiped with a diagonal length of 3 is [ boxed{18} ]

answer:To find the total number of wheels, we need to multiply the number of each type of cycle by the number of wheels they have and then add all the results together. For bicycles, there are 3 bicycles with 2 wheels each, so that's 3 * 2 = 6 wheels. For tricycles, there are 4 tricycles with 3 wheels each, so that's 4 * 3 = 12 wheels. For unicycles, there are 7 unicycles with 1 wheel each, so that's 7 * 1 = 7 wheels. Now, we add all the wheels together: 6 (bicycle wheels) + 12 (tricycle wheels) + 7 (unicycle wheels) = boxed{25} wheels in total.

answer:To determine the position of the point (P) on the line segment (AB) with respect to points (A) and (B), we will use the concept of ratios and apply specific theorems, such as Thales' theorem and Ceva's theorem. Here's the step-by-step detailed solution: 1. **Identify Key Points**: - Let ( triangle ABC ) be a triangle. - ( M ) is the midpoint of ( BC ), so ( MB = MC ). - ( N ) is the point on ( AC ) such that ( dfrac{NC}{NA} = dfrac{1}{2} ) (i.e., ( N ) divides ( AC ) in the ratio ( 1:2 )). - Line ( MN ) intersects ( AB ) at point ( P ). 2. **Use the Ratio Given**: - For point ( P ) dividing ( AB ), we need to determine the ratio ( dfrac{PA}{PB} ). 3. **Apply Ceva's Theorem** in the context of ratios: - According to Ceva's Theorem, the line segments are concurrent if: [ dfrac{MB}{MC} cdot dfrac{NC}{NA} cdot dfrac{PA}{PB} = 1 ] - Given: [ dfrac{MB}{MC} = 1 quad (text{since } M text{ is the midpoint of } BC) ] - And: [ dfrac{NC}{NA} = dfrac{1}{2} ] 4. **Substitute the Known Values**: - Substitute the known values into the equation from Ceva's Theorem: [ 1 cdot dfrac{1}{2} cdot dfrac{PA}{PB} = 1 ] 5. **Solve for the Ratio ( dfrac{PA}{PB} )**: - Simplify and solve the equation: [ dfrac{1}{2} cdot dfrac{PA}{PB} = 1 ] - Multiply both sides by 2 to isolate ( dfrac{PA}{PB} ): [ dfrac{PA}{PB} = 2 ] 6. **Interpret the Result**: - The ratio ( dfrac{PA}{PB} = 2 ) implies that point ( P ) divides ( AB ) in the ratio 2:1. - Practically, this means that ( P ) is closer to ( B ) than ( A ), and ( P ) is twice as far from ( B ) as it is from ( A ). # Conclusion Thus, point ( P ) divides the segment ( AB ) such that the distance from ( A ) to ( P ) is twice the distance from ( B ) to ( P ). This means ( P ) is the point on ( AB ) such that ( PA:PB = 2:1 ). [ boxed{frac{PA}{PB} = 2} ]