answer:Solution. From the recursive definition of the sequence, we getx_{n+1}-1=frac{1}{sqrt[10]{2}}left(x_{n}-1right), n in mathbb{N}We define the sequence y_{n}=x_{n}-1, n in mathbb{N} where y_{1}=1 and y_{n+1}=frac{1}{sqrt[10]{2}} y_{n}, n in mathbb{N}. Therefore, left{y_{n}right} is a geometric progression with the first term y_{1}=1 and common ratio frac{1}{sqrt[10]{2}}. Hence, y_{n+1}=y_{1}left(frac{1}{sqrt[10]{2}}right)^{n}, n in mathbb{N}. We have y_{1001}=frac{1}{2^{100}} and x_{n}=y_{n}+1, n in mathbb{N}, from which we getx_{1001}=1+frac{1}{2^{100}}=1+frac{5^{100}}{10^{100}}=1+underbrace{0.00 ldots 01}_{100 text { zeros }} cdot 5^{100}Thus, the first two digits of the number x_{1001} are 1 and 0, and the last two digits are 2 and 5.

answer:Answer: K=20^{2} / 4=100. In case of a 4 N times 4 M board, the answer is K=4 N M.Solution. We show two strategies, one for Horst to place at least 100 knights, and another strategy for Queenie that prevents Horst from putting more than 100 knights on the board.A strategy for Horst: Put knights only on black squares, until all black squares get occupied.Colour the squares of the board black and white in the usual way, such that the white and black squares alternate, and let Horst put his knights on black squares as long as it is possible. Two knights on squares of the same colour never attack each other. The number of black squares is 20^{2} / 2=200. The two players occupy the squares in turn, so Horst will surely find empty black squares in his first 100 steps.A strategy for Queenie: Group the squares into cycles of length 4, and after each step of Horst, occupy the opposite square in the same cycle.Consider the squares of the board as vertices of a graph; let two squares be connected if two knights on those squares would attack each other. Notice that in a 4 times 4 board the squares can be grouped into 4 cycles of length 4, as shown in Figure 1. Divide the board into parts of size 4 times 4, and perform the same grouping in every part; this way we arrange the 400 squares of the board into 100 cycles (Figure 2).The strategy of Queenie can be as follows: Whenever Horst puts a new knight to a certain square A, which is part of some cycle A-B-C-D-A, let Queenie put her queen on the opposite square C in that cycle (Figure 3). From this point, Horst cannot put any knight on A or C because those squares are already occupied, neither on B or D because those squares are attacked by the knight standing on A. Hence, Horst can put at most one knight on each cycle, that is at most 100 knights in total.Comment 1. Queenie's strategy can be prescribed by a simple rule: divide the board into 4 times 4 parts; whenever Horst puts a knight in a part P, Queenie reflects that square about the centre of P and puts her queen on the reflected square.Comment 2. The result remains the same if Queenie moves first. In the first turn, she may put her first queen arbitrarily. Later, if she has to put her next queen on a square that already contains a queen, she may move arbitrarily again.

answer:5. The inequality can be written as 1+frac{7}{8}>1+frac{k}{n}>1+frac{6}{7}. Or frac{98}{112}>frac{k}{n}>frac{96}{112}.If n=112, then the unique value of k is 97.Assuming n>112, then frac{98 n}{112 n}>frac{112 k}{112 n}>frac{96 n}{112 n}.Between 96 n and 98 n, there are at least two numbers that are multiples of 112, in which case, the value of k is not unique. Therefore, the maximum value of n is 112.

answer:6. A.Since frac{3}{4}cos frac{3}{4} > cos frac{pi}{4} = frac{sqrt{2}}{2}.Then a < b < c. Hence angle A < angle B < angle C.

answer:Answer:begin{tabular}{|l|}hline 0.0025632817 hlineend{tabular}

answer:1. frac{text { HuaCup }}{text { Shao } times text { Jun }+ text { JinTan }+ text { Lun } times text { Shu }}=15In the equation above, different Chinese characters represent different digits from 1-9. When the three-digit number “HuaCup” reaches its maximum value, please write down one way to fill in the numbers that makes the equation true.【Analysis】First, analyze that since the value of the fraction is 15, the numerator must be a multiple of 15, and the three-digit number must be the largest possible, so this three-digit number is 975. Continue reasoning to find the solution.【Solution】Solution: According to the problem:(1) Since the value of the fraction is 15, the numerator must be a multiple of 15, and the three-digit number must be the largest possible, so this three-digit number is 975.(2) Therefore, 975 div 15=65, so the denominator value is 65, and the remaining digits are 1,2,3,4,6,8.(3) If 6 times 8=48, the sum of the other digits is 17, which does not work. If 4 times 8=32, the sum of the other digits is 33, which does not work. If 3 times 8=24, the sum of the other digits is 41, which also does not work.2 times 8+43+1 times 6=65 satisfies the condition.Therefore, the answer is: 975