answer:1. **First Solution (Recurrence):** - Define ( a_{k,c} = |{mathbf{x} in mathbb{F}_5^k : mathbf{x} cdot mathbf{x} = c}| ). This represents the number of sequences of length ( k ) over ( mathbb{F}_5 ) such that the sum of the squares of the elements is ( c ). - We claim that: [ a_{k,c} = a_{k-1,c} + 2a_{k-1,c-1} + 2a_{k-1,c+2} ] This follows by considering the last component of ( mathbf{x} ). The possible values for the last component ( x_k ) are ( 0, 1, 2, 3, 4 ), and their squares modulo 5 are ( 0, 1, 4, 4, 1 ) respectively. Thus, the recurrence relation accounts for these possibilities. - Using the recurrence relation, we can derive: [ a_{k,c} = 9a_{k-2,c} + 4a_{k-2,c-1} + 4a_{k-2,c-2} + 4a_{k-2,c+1} + 4a_{k-2,c+2} = 5a_{k-2,c} + 4 cdot 5^{k-2} ] because all ( 5^{k-2} ) vectors in ( mathbb{F}_5^{k-2} ) are counted in some ( a_{k-2,c} ). - Since ( a_{0,0} = 1 ), we can inductively show that: [ a_{2n,0} = 5^{2n-1} + 5^n - 5^{n-1} ] 2. **Second Solution (Fourier analysis/Roots of unity):** - Define the polynomial ( P(x) = sum_{a=0}^4 x^{a^2} ). This polynomial captures the possible values of ( x_i^2 ) for ( x_i in {0, 1, 2, 3, 4} ). - The coefficient of ( x^m ) in ( P(x)^{2n} ) is the number of solutions to ( x_1^2 + cdots + x_{2n}^2 = m ) with each ( x_i in {0, 1, 2, 3, 4} ). - To count the total number such that ( x_1^2 + cdots + x_{2n}^2 ) is divisible by 5, we apply the inverse discrete Fourier transform (roots of unity method). Let ( omega = e^{i cdot frac{2pi}{5}} ), then the desired count is: [ frac{1}{5} sum_{k=0}^4 P(omega^k)^{2n} ] - Calculating ( P(omega^k) ): [ P(omega^k) = 1 + 2omega^k + 2omega^{4k} ] - Therefore: [ frac{1}{5} sum_{k=0}^4 P(omega^k)^{2n} = frac{1}{5} left( 5^{2n} + 2(1 + 2omega + 2omega^4)^{2n} + 2(1 + 2omega^2 + 2omega^3)^{2n} right) ] - Simplifying further: [ = 5^{2n-1} + frac{2}{5} (1 + 4 cos frac{2pi}{5})^{2n} + frac{2}{5} (1 + 4 cos frac{4pi}{5})^{2n} ] [ = 5^{2n-1} + frac{2}{5} cdot 5^n + frac{2}{5} cdot 5^n ] [ = 5^{2n-1} + 4 cdot 5^{n-1} ] The final answer is (boxed{5^{2n-1} + 4 cdot 5^{n-1}})

answer:Let the first term of the geometric sequence be a_1, and the common ratio be q, then S_3 = a_1 + a_1q + a_1q^2, S_2 = a_1 + a_1q. Since S_3 : S_2 = 3 : 2, therefore dfrac{a_1 + a_1q + a_1q^2}{a_1 + a_1q} = dfrac{3}{2}, Since a_1 neq 0, therefore dfrac{1 + q + q^2}{1 + q} = dfrac{3}{2}. Solving this, we get q = 1 or q = -dfrac{1}{2}. Therefore, the correct choice is boxed{text{D}}. This problem examines the concept and properties of geometric sequences, including the sum of the first n terms. It is a basic question.

answer:Part 1: **Problem:** Show that the points K, Z, E, and F are concyclic. 1. **Hexagon Configuration and Pascal's Theorem:** The hexagon NMPFDE is considered, and by Pascal's theorem, we know that the intersection points of opposite sides are collinear. Hence, the points Y = (NM) cap (FD), X = (MP) cap (DE), and A = (PF) cap (EN) must be collinear. 2. **Intersection of Lines and Menelaus Theorem:** Let Q be the intersection of lines (EF) and (AY). Applying Menelaus' theorem to the triangle DEF with transversal QXY, we have: frac{QE}{QF} cdot frac{YF}{YD} cdot frac{XD}{XE} = 1 3. **Applying Thales' Theorem:** From Thales’ theorem: frac{YF}{YD} = frac{DF}{YD} + 1 = frac{DB}{DM} + 1 = frac{MB}{MD} And, frac{XD}{XE} = frac{MD}{MC} = frac{MD}{MB} 4. **Equating and Simplicity:** Thus: frac{QE}{QF} = 1 This implies that Q is the midpoint of [EF]. 5. **Symmedians and Isogonals:** The indirect similarity centered at A mapping E to B and F to C will map Q to M. Therefore, lines (AM) and (AQ) are isogonal conjugates with respect to lines (AB) and (AC). This shows that line (AQ) is the A-symmedian. 6. **Harmonic Conjugates:** Thus, points A, Z, C, and B form a harmonic division. 7. **Point U and Concurrency:** Let U be the intersection point of lines (ZK) and (BC). It's known that K is on the circumcircle since it is the symmetric point of the orthocenter H with respect to [BC]. Projecting from point K onto line (BC): -1 = (A, Z, B, C) = (D, U, B, C) 8. **Proof using Harmonic Division:** Here, the point U^prime intersection of (EF) and (BC) satisfies that U^prime, D, B, and C form a harmonic set. Therefore, U = U^prime proving that lines (EF), (ZK), and (BC) concur at U. 9. **Power of a Point:** Thus, using the power of a point: UK cdot UZ = UB cdot UC = UE cdot UF Consequently, by the converse of the cyclic quadrilateral property, points E, F, K, and Z are concyclic. Conclusion for Part 1: blacksquare Part 2: **Problem:** Show that the lines (BS), (CT), and (XY) are concurrent. 1. **Pascal's Theorem and Configuration:** By Pascal's theorem applied to hexagon KTCABS, the points F = (KT) cap (AB), (CT) cap (BS), and E = (AC) cap (KS) are collinear. The goal is to show the intersection of lines (BS) and (CT) is point Q. 2. **Cyclic Quadrilateral Properties:** The cyclic quadrilaterals CEFB and HDFB yield the relationship: angle QEC = 180^circ - angle FBC = 180^circ - angle DBF = angle KHF 3. **Similarity and Ratio Argument:** Furthermore: frac{HF}{HK} = frac{1}{2} cdot frac{HF}{HD} = frac{1}{2} cdot frac{HF}{HB} cdot frac{HB}{HD} = frac{1}{2} cdot frac{HF}{HB} cdot frac{BC}{EC} = frac{1}{2} cdot frac{EF}{EC} = frac{EQ}{EC} 4. **Similarity of Triangles:** Hence, triangles QEC and FHK are similar. Thus: angle ECQ = angle HKF = angle AKT = angle ECT Therefore, points C, Q, and T are collinear. Similarly, points B, Q, and S are collinear. 5. **Concurrency Proof:** Thus, lines (BS) and (CT) intersect at a common point on line (XY). Conclusion for Part 2: blacksquare

answer:1. **Case Analysis for ( n mod 3 = 0 ):** When ( n equiv 0 pmod{3} ), suppose ( n = 3k ). Consider the total number of small squares in the ( n times n ) grid, which is ( n^2 - 1 ). We can see: [ 3 mid (n^2 - 1) ] which implies that ensuring ( n^2 - 1 ) can be covered by ( mathrm{L} )-shaped pieces (each covering 3 squares) does not work since ( n^2 - 1 ) is a multiple of 3. Therefore, it is not possible to use L-shapes to cover the whole grid minus one square. 2. **Specific Cases for ( n = 4 ):** When ( n = 4 ), consider removing one square from the top-left ( 2 times 2 ) subgrid. This leaves one ( 2 times 2 ) subgrid missing only one square, transforming into an L-shape. Then, it is possible to cut the remaining grid into L-shaped pieces. This scenario works as described. ( blacksquare ) 3. **Specific Cases for ( n = 5 ):** When ( n = 5 ), remove one square from the first row and the fourth column. As a result: [ begin{aligned} &text{Two ( 2 times 3 ) rectangles will remain (let's call them } A text{ and } B). &text{A and B are each divided into two L-shapes.} end{aligned} ] However, the remaining ( 3 times 3 ) section cannot be divided into L-shapes. Thus, it does not meet the requirement. ( blacksquare ) 4. **Specific Cases for ( n = 7 ):** Assume the removed square is located within one of the four ( 2 times 2 ) subgrids in the upper left corner ( I, II, III, ) or ( IV ). We need to: [ text{Consider only the subgrids } I, II, III text{ as shown in the diagram.} ] By symmetry, this scenario is equivalent to ( n = 4 ). Therefore, the analysis for ( n = 7 ) works similar to ( n = 4 ). ( blacksquare ) 5. **Specific Cases for ( n = 8 ):** Similar to ( n = 4 ): [ text{Removing one square from the top-left ( 2 times 2 ) subgrid allows the remaining to be partitioned completely into L-shaped pieces.} ] Therefore, there is a valid solution. ( blacksquare ) 6. **General Cases for ( n geq 10 ) with ( n = 3k pm 1 ):** When ( n = 10 ): [ text{Similar to before, remove one grid from top-left of an } 8 times 8 text{ subgrid. This ensures the grid can be partitioned into L-shapes.} ] When ( n = 11 ): [ text{Remove one square within a } 7 times 7 text{ subgrid, making it similar to the previously handled cases.} ] Generally, if ( n equiv 1 pmod{3} ) or ( n equiv 2 pmod{3} ): [ text{Extend from the cases } n = 10 text{ and } n = 11 text{ respectively, verifying possible partitions into L-shapes.} ] **Conclusion:** All ( n(n geq 3) ) fulfilling the requirement are: [ boxed{n neq 3k, n geq 3} ]