answer:Similar to the original proof, we have: sum_{i=1}^m frac{1}{3i-2}=frac 1 3sum_{i=1}^mfrac{3m}{(3i-2)(3(m+1-i)-2)} So we need to prove that the numerator of sum_{i=1}^mfrac 1{(3i-2)(3m-3i+1)} is divisible and at most divisible by 3^q. Note that (3i-2)(3m-3i+1) equiv -(3i-2)^2pmod{3^{q+1}}. Divide 2,5,cdots,3m-1 into m/3^q subsets of consecutive 3^q numbers. Then each subset consists of exactly all the units of the ring mathbb Z/3^{q+1}mathbb Z. Using the identity sum_{i=1}^m i^2=m(m+1)(2m+1)/6, we can show that the numerator is divisible and at most divisible by 3^{3q}.

answer:The null space of the matrix is the set of all vectors v=left( begin{array}{c} x_1 x_2 x_3 end{array} right) such that M.v=0, where M is the given matrix. To find the null space, we can reduce the matrix to row echelon form: left( begin{array}{ccc} 2 & -5 & 1 -5 & -3 & 7 3 & 2 & 1 end{array} right) rightarrow left( begin{array}{ccc} 1 & 0 & 0 0 & 1 & 0 0 & 0 & 1 end{array} right). The free variables in the null space correspond to the columns in the row echelon form which have no pivot. In this case, there are no free variables, so the only value of v that would make M.v=0 is the zero vector left( begin{array}{c} 0 0 0 end{array} right). Therefore, the null space is the singleton set containing only the zero vector: {, (0,0,0), }.

answer:The solution to the system of equations is x = 17, y = -19, and z = -36.

answer:Perimeter: 2.03 Type: Convex Interior Angles: {0.47,2.85,1.18,1.64} (Note: These angles have been corrected for accuracy.) Area: 0.11 (The original area was underestimated. The revised area is calculated using the shoelace formula or other suitable methods for finding the area of a convex quadrilateral.)