Awesome thinking - thank you! The (Historically) First(?) For instance, this one: $$ 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1 $$. @Omega: I think of it as standard. Module 4: Mathematical Induction Theme 1: Principle of Mathematical Induction Mathematical induction is used to prove statements about natural numbers. Is an IP68 rating sufficient to protect a phone during a 12 hour ride in heavy rain? Page 5. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. \)\( \begin{aligned} \require{color}\text{LHS } &= (2k+2)! How to use mathematical induction with inequalities? 440 0 obj<>stream The statement P0 says that p0 = 1 = cos(0 ) = 1, which is true.The statement P1 says that p1 = cos = cos(1 ), which is true. It's similar to proof of equality. > 2^{k+1} \big[(k+1)!\big]^2 \text{ for } \ge 2.\), Absolute Value Algebra Arithmetic Mean Arithmetic Sequence Binomial Expansion Binomial Theorem Chain Rule Circle Geometry Common Difference Common Ratio Compound Interest Cyclic Quadrilateral Differentiation Discriminant Double-Angle Formula Equation Exponent Exponential Function Factorials Functions Geometric Mean Geometric Sequence Geometric Series Inequality Integration Integration by Parts Kinematics Logarithm Logarithmic Functions Mathematical Induction Polynomial Probability Product Rule Proof Quadratic Quotient Rule Rational Functions Sequence Sketching Graphs Surds Transformation Trigonometric Functions Trigonometric Properties VCE Mathematics Volume, Your email address will not be published. = 16 \\\text{RHS } &= 2^2 \times (2!) The inequality holds for $n = 1$. \\&> 2(k+1)(2k+1)2^k (k! This part is not quite mechanical, but the $1+\cdots+\frac{1}{k}$ part of $(2)$ is kind of saying "use what $(1)$ tells you about me." $$ So Prove that \( (2n)! I've been using mathematical induction to prove propositions like this: $$1 + 3 + 5 + \cdots + (2n-1) = n^2$$ Which is an equality. Now comes the nontrivial part (though not hard in this case), where we need to somehow get $(n+1)/2+1$. This was the only part that was a bit different from the inductions you have done. ($2^n
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