answer:**Objective:** We are to determine if there exists a function (f: mathbb{N}^* to mathbb{N}^*) that is strictly increasing and satisfies ( f(1) = 2 ) and for all ( n in mathbb{N}^* ): [ f(f(n)) = f(n) + n. ] **Approach:** Construct such a function by induction. **Steps:** 1. **Base case:** Define ( f(1) = 2 ). 2. **Induction hypothesis:** Assume we have defined ( f ) for all ( 1 leq k leq n ) such that ( f ) is strictly increasing and the given functional equation holds for all ( k leq n ). 3. **Inductive step:** Define ( f(n+1) ). We need ( f ) to be strictly increasing, so we begin by ensuring ( f(n+1) > f(n) ). We also need to satisfy the functional equation for ( n+1 ): [ f(f(n+1)) = f(n+1) + (n + 1). ] Let's denote a potential candidate for ( f(n+1) ) by ( a ), where ( a ) is such that ( f(n) < a ) and ( a ) adheres to the equation above. 4. **Constructing ( f(n+1) ):** - Define a helper function ( g ): [ g(n + 1) = max{ k in {1, ldots, n} mid f(k) leq n + 1 }. ] Since ( g(n + 1) ) ensures the maximum ( k ) with ( f(k) leq n + 1 ), and we know ( g(n + 1) ) exists because ( f(1) = 2 leq n+1 ). - Then set: [ f(n + 1) = g(n + 1) + n + 1. ] - By construction, this definition makes ( f ) strictly increasing because ( g(n + 1) geq g(n) implies f(n + 1) > f(n) ). 5. **Verification:** Now, verify if the functional equation holds: [ f(f(n + 1)) = f(g(n + 1) + n + 1). ] Given that ( g(f(n + 1)) = n + 1 ), we get: [ f(f(n + 1)) = f(n + 1) + (n + 1). ] - Check the base case consistency: - For ( f(1) = 2 ), ( f(f(1)) = f(2) = g(2) + 2 = 1 + 2 = 3 ), we have: [ f(f(1)) = 3 = 2 + 1 = f(1) + 1. ] - Inductive case consistency: - Assume ( f(n) ) is accurately defined. By construction, ( f(n+1) ) with ( g(f(n+1)) = n+1 ): [ f(f(n+1)) = f(n+1) + (n + 1). ] **Conclusion:** Thus, there exists a strictly increasing function ( f: mathbb{N}^* to mathbb{N}^* ) such that ( f(1) = 2 ) and ( f(f(n)) = f(n) + n ) for all ( n ). Therefore, the answer is: [ boxed{text{Yes}} ]

answer:The geometric sequence given has a first term a_0 = frac{1}{3} and a common ratio r = frac{1}{3}. The number of terms, n, is 8. The sum S_n of the first ( n ) terms of a geometric sequence can be calculated using the formula: [ S_n = a_0 frac{1-r^n}{1-r} ] Substituting the known values: [ S_8 = frac{frac{1}{3}(1-left(frac{1}{3}right)^8)}{1-frac{1}{3}} = frac{frac{1}{3}(1-frac{1}{6561})}{frac{2}{3}} = frac{frac{1}{3} times frac{6560}{6561}}{frac{2}{3}} = frac{6560}{19683} ] Thus, the sum of the first eight terms of the sequence is: [ boxed{frac{6560}{19683}} ]

answer:To solve, 101111_{(2)} = 1 times 2^5 + 0 times 2^4 + 1 times 2^3 + 1 times 2^2 + 1 times 2^1 + 1 times 2^0 = 47. Therefore, the correct option is boxed{A}. **Analysis:** By using 101111_{(2)} = 1 times 2^5 + 0 times 2^4 + 1 times 2^3 + 1 times 2^2 + 1 times 2^1 + 1 times 2^0, we can easily find the answer.

answer:First, let's calculate how much Dave can bench press. Since Dave can bench press three times his body weight and he weighs 175 pounds: Dave's bench press = 3 * 175 = 525 pounds Mark can bench press 55 pounds, and we know that Craig can bench press 50 pounds more than Mark: Craig's bench press = Mark's bench press + 50 = 55 + 50 = 105 pounds Now, to find the percentage of the amount Craig can bench press compared to Dave's bench press, we use the formula: Percentage = (Craig's bench press / Dave's bench press) * 100 Percentage = (105 / 525) * 100 Percentage = (1 / 5) * 100 Percentage = 0.2 * 100 Percentage = 20% So, Craig can bench press boxed{20%} of the amount Dave can bench press.