answer:1. **Understand the direct expression**: [2^{341}+1 = 100ldots01_2 ] (340 zeros) [2^{31}+1 = 100ldots01_2 ] (30 zeros) 2. **Express (frac{1}{2^{31}+1}) as a geometric series**: [ frac{1}{2^{31}+1} = 2^0 - 2^{31} + 2^{62} - 2^{93} + ldots ] 3. **Form the multiplication to simulate reduction**: [ (2^{341} + 1)(2^0 - 2^{31} + 2^{62} - ldots) ] Consider the sequence and cancellation in pairs ((2^{341}cdot 2^{0}, 2^{341}cdot2^{31}, ldots)) reducing by ((2^{31}, 2^{62}, ldots)) results in the need to monitor each power's residue. 4. **Detailed construction and cancellation**: Each subtraction (2^{n+31} - 2^n) leads to replacement with (2^{n+30} + 2^{n+29} + ldots + 2^n). This introduces 31 new terms for each difference. 5. **Count intervals of contribution**: Noting that the highest non-cancelable term's power is (310 = 341 - 31), we count the differences at intervals up to (310). 6. **Summarize the count**: Each interval (x = 0, 31, ldots, 310) contributes 31 terms. Therefore, with 11 intervals, (31 times 11 = 341) terms. 7. **Result**: [ 341 text{ terms} ] The final answer is boxed{D) 341}

answer:1. **Establish the Given Elements**: Let ( l ) be the given straight line and ( h ) be the given distance. 2. **Creating Perpendicular Line**: Through an arbitrary point ( M ) that lies on the given line ( l ), draw a straight line that is perpendicular to ( l ). 3. **Selecting Points at Distance ( h )**: On this perpendicular line, select two points ( A ) and ( B ), which are located on opposite sides of the line ( l ) such that ( MA = MB = h ). 4. **Identifying Parallel Lines**: Let's denote the lines passing through points ( A ) and ( B ), respectively, that are parallel to line ( l ) as ( l_1 ) and ( l_2 ). 5. **Proof for Points on ( l_1 ) and ( l_2 )**: We need to prove that these lines ( l_1 ) and ( l_2 ) configure the geometric place of points that are at a given distance ( h ) from the line ( l ). - Assume an arbitrary point ( X ) other than ( A ) lies on line ( l_1 ). - Drop a perpendicular from point ( X ) to line ( l ), and call the foot of the perpendicular ( X' ). Construct the rectangle ( AXX'M ). - By the property of the rectangle, the length of the perpendicular ( XX' ) is equal to ( AM = h ). Thus, the point ( X ) is at a distance ( h ) from line ( l ). 6. **Generalization to Any Point on ( l_1 ) and ( l_2 )**: - Similarly, perform the proof for any point lying on line ( l_2 ) and verify that the perpendicular distance to line ( l ) is ( h ). 7. **Validation for Points Given ( h )**: - Suppose now a point ( Y ) is given at a distance ( h ) away from line ( l ). Assume ( Y ) and ( A ) are on the same side of line ( l ). - Drop a perpendicular from ( Y ) to ( l ), denoting the foot of the perpendicular as ( Y' ). Construct the rectangle ( AYY'M ). - The geometry implies that ( AY | MY' ), meaning ( AY | l ). Since there can be only one line parallel to ( l ) passing through a non-collinear point, the line ( AY ) coincides with ( l_1 ). Therefore, point ( Y ) lies on ( l_1 ). 8. **Alternative Positioning**: - If ( Y ) and ( A ) are on opposite sides of line ( l ), similarly demonstrate that point ( Y ) lies on ( l_2 ). # Conclusion: The geometric place of points that are equidistant from a given straight line ( l ) is two parallel straight lines ( l_1 ) and ( l_2 ), each at a perpendicular distance ( h ) from ( l ). [ boxed{l_1 text{ and } l_2} ]

answer:Solution: y'=x^2, when x=1, y'=1, thus the slope angle of the tangent line is 45^circ. Therefore, the answer is 45^circ. By understanding the geometric meaning of the derivative, we calculate the derivative of the function f(x) at x=1 to find the slope of the tangent line, and thus determine the slope angle of the tangent line. This question mainly examines the ability to use derivatives to study the equation of the tangent line at a certain point on a curve, focusing on computational problem-solving skills, and is considered a medium-level question. The final answer is boxed{45^circ}.

answer:Define a new sequence (b_n) such that a_n = 2^n b_n for each n. Then the recurrence for (a_n) can be rewritten as: [ 2^{n+1} b_{n+1} = frac{10}{6} cdot 2^n b_n + frac{8}{6} sqrt{4^n - 4^n b_n^2} ] [ 2^{n+1} b_{n+1} = frac{10}{6} cdot 2^n b_n + frac{8}{6} cdot 2^n sqrt{1 - b_n^2} ] Divide both sides by (2^{n+1}) to simplify: [ b_{n+1} = frac{5}{3} b_n + frac{4}{3} sqrt{1 - b_n^2} ] Compute values: [ b_1 = frac{4}{3} ] [ b_2 = frac{5}{3} cdot frac{4}{3} + frac{4}{3} sqrt{1 - left(frac{4}{3}right)^2} ] [ b_2 = frac{20}{9} + frac{4}{3} cdot frac{1}{3} = frac{20}{9} + frac{4}{9} = frac{24}{9} = frac{8}{3} ] Perform similar calculations for b_3, b_4, b_5. Assume no cycles initially: [ b_3 = frac{5}{3} cdot frac{8}{3} + frac{4}{3} sqrt{1 - left(frac{8}{3}right)^2} ] [ b_3 = frac{40}{9} + frac{4}{3} cdot 0 = frac{40}{9} ] [ b_4 = frac{5}{3} cdot frac{40}{9} = frac{200}{27} ] [ b_5 = frac{5}{3} cdot frac{200}{27} = frac{1000}{81} ] Thus a_5 = 2^5 cdot frac{1000}{81} = 32 cdot frac{1000}{81} = boxed{frac{32000}{81}}.