answer:Solution. Let's denote the sum by S_{n}:S_{n}=1^{3}+3^{3}+5^{3}+ldots(2 n-1)^{3}We need to find an expression for the general term S_{n} that would allow us to compute the sum S_{n} for any n without raising all numbers from 1 to 2 n-1 to the third power.Consider the first differences of the sequence defined by formula (1). The general form of the first differences isbegin{gathered}d_{n}=S_{n+1}-S_{n}=left(1^{3}+3^{3}+5^{3}+ldots+(2 n-1)^{3}+(2 n+1)^{3}right)- -left(1^{3}+3^{3}+5^{3}+ldots+(2 n-1)^{3}right)=(2 n+1)^{3}end{gathered}- a polynomial of the third degree. Therefore, the general term of the sequence defined by formula (1) can be represented as a polynomial of the fourth order:S_{n}=a x^{4}+b x^{3}+c x^{1}+d x+l .To determine the coefficients a, b, c, d and l, substitute the values of S_{n} for n=1,2,3,4,5 into this formula, then we get the system of equations:left{begin{aligned}a+b+c+d+l & =1 16 a+8 b+4 c+2 d+l & =28 81 a+27 b+9 c+3 d+l & =153 256 a+64 b+16 c+4 d+l & =496 625 a+125 b+25 c+5 d+l & =1225end{aligned}right.The solutions to this system are: a=2, b=0, c=-1, d=l=0.Therefore, S_{n}=2 n^{4}-n^{2}=n^{2}left(2 n^{2}-1right).

answer:(1) Since a_4 = a_2a_3, we have a_4 = a_1a_4.As a_4 neq 0, it follows that a_1 = 1.Given S_3 = a_1 + a_2 + a_3 = 1 + q + q^2 = 13 and q > 0, we solve for q and find q = 3.Therefore, a_n = a_1q^{n-1} = 3^{n-1}.(2) Since b_n = log_3 3^{n-1} + n = 2n - 1,we have frac{1}{b_n b_{n+1}} = frac{1}{(2n-1)(2n+1)} = frac{1}{2}(frac{1}{2n-1} - frac{1}{2n+1}).Hence, T_n = frac{1}{2}[(1 - frac{1}{3}) + (frac{1}{3} - frac{1}{5}) + dots + (frac{1}{2n-1} - frac{1}{2n+1})]= frac{1}{2} times (1 - frac{1}{2n+1}) = boxed{frac{n}{2n+1}}.

answer:Thenumberproblemsolution First, prove that a, b, c, d are all not less than 2, and are pairwise coprime.This is because, according to the problem, we have b c d = k a + 1. Clearly, a geqslant 2, otherwise, if a=1, then b c d divided by a would have a remainder of 0, which contradicts the problem statement.At the same time, a is coprime with any of b, c, d.Similarly, b geqslant 2, c geqslant 2, d geqslant 2, and a, b, c, d are pairwise coprime.Without loss of generality, assume 2 leqslant a leqslant 5, d geqslant 7.Equation (1) can be rewritten as1 + frac{1}{6 c d} = frac{1}{2} + frac{1}{3} + frac{1}{c} + frac{1}{d}, thusfrac{1}{c} + frac{1}{d} = frac{1}{6} + frac{1}{6 c d} < frac{1}{6}, so c < 12.From a=2, we know c is an odd number, hence c leqslant 11.From equation (3), we can derive d = 6 + frac{35}{c-6}.Since d geqslant 7 and d is a positive integer, frac{35}{c-6} must be a positive integer. Considering c leqslant 11, we have 1 leqslant c-6 leqslant 5, thus c-6=1 or c-6=5, i.e.,c=7 or c=11.When c=7, from equation (4) we get d=41,When c=11, from equation (4) we get d=13.Therefore, the four numbers (a, b, c, d) = (2,3,7,11) or (2,3,11,13).

answer:## Solutionbegin{aligned}& lim _{n rightarrow infty} frac{(n+2)^{4}-(n-2)^{4}}{(n+5)^{2}+(n-5)^{2}}=lim _{n rightarrow infty} frac{left((n+2)^{2}-(n-2)^{2}right) cdotleft((n+2)^{2}+(n-2)^{2}right)}{(n+5)^{2}+(n-5)^{2}}= & =lim _{n rightarrow infty} frac{left(n^{2}+4 n+4-n^{2}+4 n-4right)left(n^{2}+4 n+4+n^{2}-4 n+4right)}{(n+5)^{2}+(n-5)^{2}}= & lim _{n rightarrow infty} frac{frac{1}{n^{2}} 8 nleft(2 n^{2}+8right)}{frac{1}{n^{2}}left((n+5)^{2}+(n-5)^{2}right)}=lim _{n rightarrow infty} frac{16 nleft(1+frac{4}{n^{2}}right)}{left(1+frac{5}{n}right)^{2}+left(1-frac{5}{n}right)^{2}}=+inftyend{aligned}## Problem Kuznetsov Limits 3-23

answer:IV. angle A l^{prime} M=45^{circ}. Hint:Solution 1 Draw a perpendicular to BC through C, and intercept CD=AM=BC on it, then connect DN, DA, DM. It can be proven that triangle DCN cong triangle CBM. Thus, triangle ADN is an isosceles right triangle.Solution 2 Outside triangle ABC, draw a perpendicular to AB through M, and intercept ME=MB=CN on it, then connect AE, NE, MN. It is easy to prove that triangle CMN cong triangle EMN, hence CM parallel EN. Through calculation, it can still be deduced that triangle AEN is an isosceles right triangle.Solution 3 Draw a perpendicular to AB through A, and intercept AK=CN=MB on it, then connect KM, KC. It is easy to know that triangle KMC is an isosceles right triangle. Also, triangle AKC cong triangle CNA, hence KC parallel AN.

answer:To determine the range of values for a, we first analyze the inequalities defining sets A and B:For set A, consider the quadratic inequality x^2 - 2x + a leq 0. Factoring or completing the square might not be straightforward without knowing a. However, we can still deduce some properties from the inequality:The parabola y = x^2 - 2x + a opens upwards since the coefficient of x^2 is positive.If A is a set of real numbers defined by this inequality, the vertex of the parabola represents the minimum point on the graph, and the value of x at the vertex is given by x = frac{-(-2)}{2 cdot 1} = 1. Since the inequality is less than or equal to zero, the range of x within A must include x=1 and extend to the left and right towards the roots of the equation (if they exist).For set B, the quadratic inequality is x^2 - 3x + 2 leq 0. This can be factored into (x-1)(x-2) leq 0. The roots of the inequality are x=1 and x=2, and since it's a "less than or equal to" inequality, the solutions are in the closed interval [1,2].Now, given that B subset A, the interval [1,2] must be completely contained within the interval defined by set A. This means that the value x=2, which is the upper bound of B, must be less than or equal to the upper bound of A.Turning our attention to the quadratic equation x^2 - 2x + a = 0, associated with the boundary of A, the sum and product of its roots, which we'll call m and n, are related to the coefficients by Vieta's formulas: m + n = 2 quad text{(sum of roots)} m cdot n = a quad text{(product of roots)} Since n must be at least 2 to include the interval [1,2], and m + n = 2, we have that m leq 0. Therefore, a = m cdot n leq 0.The range of values for a is such that it supports these conditions, thus we conclude:boxed{a leq 0}.