answer:Let S = sum_{n = 0}^infty frac{F_n}{2^n}. Then, mimicking the manipulation used in the original problem: [ S = F_0 + frac{F_1}{2} + frac{F_2}{2^2} + frac{F_3}{2^3} + dotsb ] [ S = frac{1}{2} + frac{F_0}{2^2} + frac{F_1}{2^3} + frac{F_2}{2^4} + dotsb + frac{F_1}{2^2} + frac{F_2}{2^3} + frac{F_3}{2^4} + dotsb ] [ S = frac{1}{2} + frac{1}{2^2}S + frac{1}{2}S. ] Combine terms and solve for S: [ S - left(frac{1}{4} + frac{1}{2}right)S = frac{1}{2} ] [ S - frac{3}{4}S = frac{1}{2} ] [ frac{1}{4}S = frac{1}{2} ] [ S = 2 ] Thus, the sum is boxed{2}.

answer:1. **Introduction to the Radical Axis:** The radical axis of two circles is the locus of points that have equal power with respect to the two circles. For any point ( P ) on the radical axis of two circles with equations [ (x - a_1)^2 + (y - b_1)^2 = r_1^2 quad text{and} quad (x - a_2)^2 + (y - b_2)^2 = r_2^2, ] the power of ( P ) with respect to both circles is the same. Specifically, if ( P(x, y) ) is on the radical axis, we have: [ (x - a_1)^2 + (y - b_1)^2 - r_1^2 = (x - a_2)^2 + (y - b_2)^2 - r_2^2. ] Simplifying this equation, we get the equation of the radical axis as: [ 2(x(a_1 - a_2) + y(b_1 - b_2)) = r_1^2 - r_2^2 + a_2^2 - a_1^2 + b_2^2 - b_1^2. ] 2. **Circle Bundles and the Radical Axis:** Consider a bundle (or family) of circles determined by the linear combination of the equations of two given circles: [ lambda ( (x - a_1)^2 + (y - b_1)^2 - r_1^2 ) + mu ( (x - a_2)^2 + (y - b_2)^2 - r_2^2 ) = 0, ] where ( lambda, mu ) are parameters. 3. **Intersection with the Radical Axis:** To determine the relationship between the circles in this bundle and the radical axis formed by fixed points, consider the equation from above and the definition of the radical axis. - If a specific circle in the bundle intersects the radical axis at two distinct points, these points are fixed since the radical axis is invariant for all circles in the bundle. This is the elliptic bunch scenario. - If a specific circle in the bundle is tangent to the radical axis, it touches it at exactly one point—also a fixed point—forming a parabolic bunch. - If a specific circle in the bundle does not intersect the radical axis, then no points are shared between them, forming a hyperbolic bunch. 4. **Proof and Conclusion:** Using the definition and properties of the radical axis, we conclude that: - For any family of circles (circle bundle), each circle can either: 1. Intersect the radical axis at exactly two fixed points (elliptic bundle); 2. Be tangent to the radical axis at one fixed point (parabolic bundle); 3. Not intersect the radical axis at all (hyperbolic bundle). Thus, the assertion is proven. [ boxed{} ]

answer:To find the rightmost three digits of 5^{1993}, we compute the powers of 5 modulo 1000 to observe any patterns that might emerge. This approach allows us to simplify the problem significantly. Let's proceed step by step: 1. **Starting with the base case**, we have: - 5^0 equiv 1 pmod{1000}, which means 5^0 leaves a remainder of 1 when divided by 1000. - 5^1 equiv 5 pmod{1000}, indicating 5^1 leaves a remainder of 5 when divided by 1000. - 5^2 equiv 25 pmod{1000}, so 5^2 leaves a remainder of 25 when divided by 1000. - 5^3 equiv 125 pmod{1000}, which means 5^3 leaves a remainder of 125 when divided by 1000. - 5^4 equiv 625 pmod{1000}, indicating 5^4 leaves a remainder of 625 when divided by 1000. 2. **Observing a pattern**, we notice that: - 5^5 equiv 3125 equiv 125 pmod{1000}, which simplifies to 125 because 3125 - 3000 = 125, and 3000 is a multiple of 1000. This observation reveals a repeating pattern every two terms starting at the 4th term (5^4). Specifically, for n > 2 and when n is odd, we have: [5^n equiv 125 pmod{1000}] Given that 1993 is odd and greater than 2, we can directly apply this pattern to conclude that: [5^{1993} equiv 125 pmod{1000}] Therefore, the rightmost three digits of 5^{1993} are boxed{125}.

answer:1. **Total Number of Invitations Sent:** Each of the 2000 people sends out 1000 friend invitations. Therefore, the total number of invitations is calculated as: [ 2000 times 1000 = 2,000,000 ] 2. **Total Number of Pairs:** Out of the (2000) people, the number of possible pairs they can form is given by the combination formula (C(n, 2)), where (n) is the number of people: [ C(2000, 2) = frac{2000 times (2000 - 1)}{2} = frac{2000 times 1999}{2} = 1,999,000 ] 3. **Comparison of Friend Invitations to Pairs:** There are (2,000,000) invitations and (1,999,000) possible pairs. Since an invitation forms a valid pair of friends only if both people invite each other back, and there are more invitations than possible unique pairs, it indicates potential for repeated or mutual invitations within pairs. 4. **Minimum Number of Friend Pairs:** Since the number of friend pairs must be such that each friend pair has sent mutual invitations, we observe that there must be at least: [ 2,000,000 - 1,999,000 = 1,000 ] mutual invitations forming pairs. 5. **Example for Clarification:** Let's arrange the 2000 people in the form of a 2000-sided polygon. Each person sends invitations to the next 1000 people in a sequential manner (according to the direction of the polygon). According to this setup, a person will be friends with the individuals situated exactly opposite them in such a polygon. This theoretically ensures that every person has exactly one mutual friend resulting in exactly 1000 friend pairs. 6. **Conclusion:** Hence, the minimum number of friend pairs that could have formed, verified through constructive example and logical minimum checks, is: [ boxed{1000} ]