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.
ScanSolve AI Answer
Here is the proof using the Principle of Mathematical Induction. 10. Prove that n! 2^n-1 for all n 1. Let P(n) be the statement n! 2^n-1. Base Case (n=1): LHS = 1! = 1 RHS = 2^1-1 = 2^0 = 1 Since 1 1, P(1) is true. Inductive Hypothesis: Assume P(m) is true for some positive integer m. That is, assume m! 2^m-1. Inductive Step: We need to prove that P(m+1) is true, i.e., (m+1)! 2^(m+1)-1, which simplifies to (m+1)! 2^m. Consider the LHS of P(m+1): (m+1)! = (m+1) · m! By the Inductive Hypothesis, we know m! 2^m-1. Substitute this into the expression: (m+1)! (m+1) · 2^m-1 Since m 1, it follows that m+1 2. Therefore, we can write: (m+1) · 2^m-1 2 · 2^m-1 Simplify the right side: 2 · 2^m-1 = 2^1+(m-1) = 2^m Combining the inequalities, we have: (m+1)! (m+1) · 2^m-1 2^m Thus, (m+1)! 2^m. This proves P(m+1) is true. By the Principle of Mathematical Induction, P(n) is true for all n 1. The proof is complete.