This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here is the solution to the problem. Question 10: Prove that n! 2^n-1 for all n 1. We will prove this by mathematical induction. Let P(n) be the statement "n! 2^n-1". Step 1: Base Case (n=1). For n=1, we check if P(1) is true. Left Hand Side (LHS): 1! = 1. Right Hand Side (RHS): 2^1-1 = 2^0 = 1. Since 1 1, the statement P(1) is true. Step 2: Inductive Hypothesis. Assume that P(k) is true for some positive integer k 1. This means we assume k! 2^k-1. Step 3: Inductive Step. We need to prove that P(k+1) is true, i.e., (k+1)! 2^(k+1)-1, which simplifies to (k+1)! 2^k. Consider the Left Hand Side of P(k+1): (k+1)! By the definition of factorial, (k+1)! = (k+1) · k!. From our inductive hypothesis, we know that k! 2^k-1. Substitute this into the expression: (k+1)! (k+1) · 2^k-1 Now, we need to show that (k+1) · 2^k-1 2^k. We know that 2^k = 2 · 2^k-1. So, we need to show that (k+1) · 2^k-1 2 · 2^k-1. Since 2^k-1 is a positive value for k 1, we can divide both sides of the inequality by 2^k-1 without changing the direction of the inequality: k+1 2 Since k is a positive integer and k 1, it follows that k+1 1+1, so k+1 2. This inequality is true for all k 1. Therefore, we have shown that (k+1)! (k+1) · 2^k-1 2 · 2^k-1 = 2^k. Thus, (k+1)! 2^k. This proves that P(k+1) is true. Step 4: Conclusion. By the principle of mathematical induction, the statement n! 2^n-1 is true for all positive integers n 1.