answer:(1) Since f(x) = ka^x - a^{-x} (a > 0 and a neq 1) is an odd function, we have f(0) = 0, which means k - 1 = 0. Solving this, we get k = 1. So, the value of k is boxed{1}. (2) Since f(x) = a^x - a^{-x} (a > 0 and a neq 1) is an odd function, the inequality f(x+2) + f(3-2x) > 0 is equivalent to f(x+2) > -f(3-2x) = f(2x-3). Because 0 < a < 1, f(x) is a monotonically decreasing function on mathbb{R}, thus x+2 < 2x-3, which simplifies to x > 5. Therefore, the range of x is boxed{(5, +infty)}. (3) Since f(1) = frac{8}{3}, we have a - frac{1}{a} = frac{8}{3}, which leads to 3a^2 - 8a - 3 = 0. Solving this, we get a = 3 or a = -frac{1}{3} (discard this solution). Thus, g(x) = 3^{2x} + 3^{-2x} - 2m(3^x - 3^{-x}) = (3^x - 3^{-x})^2 - 2m(3^x - 3^{-x}) + 2, Let t = 3^x - 3^{-x}, Since x geq 1, we have t geq f(1) = frac{8}{3}, thus (3^x - 3^{-x})^2 - 2m(3^x - 3^{-x}) + 2 = (t-m)^2 + 2 - m^2, Since the function g(x) has a minimum value of -2 on [1, +infty), when m geq frac{8}{3}, 2 - m^2 = -2, solving this gives m = 2, which is discarded. When m < frac{8}{3}, solving (frac{8}{3})^2 - 2m times frac{8}{3} + 2 = -2 gives m = frac{25}{12}, which satisfies the condition. Therefore, m = boxed{frac{25}{12}}.

answer:Let: - ( t ) be the time it takes for the jet ski to travel from dock A to dock B). - ( r ) be the speed of the river current (and the canoe, since it drifts with the current). - ( p ) be the speed of the jet ski relative to the water. Given: - ( p = r + 10 text{ km/h} ) - Distance from dock A to dock B is 60 km. 1. **Distance Relationships:** - The effective speed of the jet ski downstream is ( p + r ). - The distance from A to B, which is 60 km, leads to: [ (p + r)t = 60 ] 2. **Time Relationship:** - The jet ski meets the canoe after traveling upstream for ( 8 - t ) hours. The distance the canoe has traveled by then is ( 8r ). - The distance traveled by the jet ski when it meets the canoe is: [ (p-r)(8-t) ] - Since these distances are the same, we have: [ 60 + (p-r)(8-t) = 8r ] 3. **Substitute and solve:** - Substitute ( p = r + 10 ) in the second equation: [ 60 + (r+10-r)(8-t) = 8r ] [ 60 + 10*(8-t) = 8r ] - From the first equation: [ (r+10+r)t = 60 ] [ 2r t + 10t = 60 ] [ t = frac{60}{2r+10} ] - Plugging ( t ) back into ( 60 + 10*(8-t) = 8r ) and solving for ( r ) and then ( t ). From the equations: - ( r = 5 text{ km/h} ) - ( t = frac{60}{2(5)+10} = 3 text{ hours} ) Conclusion: The time it took the jet ski to reach dock B is 3 hours. The final answer is boxed{textbf{B}}

answer:1. **Base Case:** First, we verify the statement for ( n = 2 ): [ 2 + h(1) = 2 + 1 = 2(1 + frac{1}{2}) = 2h(2) ] Hence, the equation ( n + h(1) + h(2) + cdots + h(n-1) = n h(n) ) holds for ( n = 2 ). 2. **Induction Hypothesis:** Assume that the equation holds for some ( n geq 2 ): [ n + h(1) + h(2) + cdots + h(n-1) = n h(n) ] 3. **Induction Step:** We need to prove that the equation holds for ( n+1 ): [ (n+1) + h(1) + h(2) + cdots + h(n-1) + h(n) = (n+1) h(n+1) ] Using the induction hypothesis: [ n + h(1) + h(2) + cdots + h(n-1) = n h(n) ] Now, add ( 1 + h(n) ) to both sides of the equation: [ n + h(1) + h(2) + cdots + h(n-1) + 1 + h(n) = n h(n) + 1 + h(n) ] Simplify the right-hand side: [ n h(n) + 1 + h(n) = n h(n) + h(n) + 1 ] [ = (n + 1) h(n) + 1 ] Recognize that: [ h(n+1) = h(n) + frac{1}{n+1} ] This gives: [ (n+1)h(n) + 1 = (n+1)left( h(n+1) - frac{1}{n+1} right) + 1 ] [ = (n+1)h(n+1) - (n+1) cdot frac{1}{n+1} + 1 ] [ = (n+1)h(n+1) - 1 + 1 ] [ = (n+1) h(n+1) ] 4. **Conclusion:** Since the equation holds for ( n=2 ) (base case) and assuming it holds for ( n ), it also holds for ( n+1 ) (induction step). By the principle of mathematical induction, the given equation is true for all integers ( n geq 2 ). Therefore, the summation formula is: [ n + h(1) + h(2) + cdots + h(n-1) = n h(n) ] (blacksquare)

answer:Solution: (1) Let n=1, we get 2a_1 - a_1 = a_1^2. That is, a_1 = a_1^2, Since a_1 neq 0, we have a_1 = 1, Let n=2, we get 2a_2 - 1 = 1 cdot (1 + a_2), solving this gives a_2 = 2, When n geqslant 2, from 2a_n - 1 = S_n, we have 2a_{n-1} - 1 = S_{n-1}, Subtracting these two equations gives 2a_n - 2a_{n-1} = a_n, that is, a_n = 2a_{n-1}, Therefore, the sequence {a_n} is a geometric sequence with the first term being 1 and the common ratio being 2, Therefore, the general formula for the sequence {a_n} is a_n = 2^{n-1}; (2) From (1), we know that na_n = n cdot 2^{n-1}, let the sum of the first n terms of the sequence {na_n} be T_n, Then T_n = 1 + 2 times 2 + 3 times 2^2 + ldots + n times 2^{n-1}, (1) 2T_n = 1 times 2 + 2 times 2^2 + 3 times 2^3 + ldots + n times 2^n, (2) Subtracting (2) from (1) gives -T_n = 1 + 2 + 2^2 + ldots + 2^{n-1} - n cdot 2^n = 2^n - 1 - n cdot 2^n, Therefore, T_n = 1 + (n-1)2^n. Thus, the answers are: (1) The general formula for the sequence {a_n} is boxed{a_n = 2^{n-1}}. (2) The sum of the first n terms of the sequence {na_n} is boxed{T_n = 1 + (n-1)2^n}.