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- https://math.libretexts.org/Courses/Monroe_Community_College/MTH_220_Discrete_Math/3%3A_Proof_Techniques/3.6%3A_Mathematical_Induction_-_An_IntroductionConsider \[P(n): \qquad n^2+n+11 \mbox{ is prime}.\] In the inductive step, we want to prove that \[P(k) \Rightarrow P(k+1) \qquad\mbox{ for ANY } k\geq1.\] The following table verifies that it is tru...Consider \[P(n): \qquad n^2+n+11 \mbox{ is prime}.\] In the inductive step, we want to prove that \[P(k) \Rightarrow P(k+1) \qquad\mbox{ for ANY } k\geq1.\] The following table verifies that it is true for \(1\leq k\leq 9\): \[\begin{array}{|*{10}{c|}} \hline n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline n^2+n+11 & 13 & 17 & 23 & 31 & 41 & 53 & 67 & 83 & 101 \\ \hline \end{array}\] Nonetheless, when \(n=10\), \(n^2+n+11=121\) is composite.
- https://math.libretexts.org/Courses/De_Anza_College/Linear_Algebra%3A_A_First_Course/08%3A_Appendices/8.02%3A_Well_Ordering_and_InductionThis page introduces summation notation and its applications, emphasizing well-ordered sets and mathematical induction. It explains how summation notation provides a concise representation of sums and...This page introduces summation notation and its applications, emphasizing well-ordered sets and mathematical induction. It explains how summation notation provides a concise representation of sums and describes the principle of well-ordering underlying induction. The section outlines the induction process, including base cases and steps, illustrated by examples that prove formulas and inequalities for all natural numbers.
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/A_Spiral_Workbook_for_Discrete_Mathematics_(Kwong)/03%3A_Proof_Techniques/3.04%3A_Mathematical_Induction_-_An_IntroductionMathematical induction can be used to prove that an identity is valid for all integers n≥1 .
- https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Barrus_and_Clark)/01%3A_Chapters/1.03%3A_Proof_by_InductionWe take the equation \[1^2+2^2+\cdots+k^2=\frac{1}{6}k(k+1)(2k+1)\nonumber \] (which is \(P(k)\), one of the equations we agreed to suppose was true in the induction hypothesis) and add \((k+1)^2\) to...We take the equation \[1^2+2^2+\cdots+k^2=\frac{1}{6}k(k+1)(2k+1)\nonumber \] (which is \(P(k)\), one of the equations we agreed to suppose was true in the induction hypothesis) and add \((k+1)^2\) to both sides to get \[ \label{indeq1} 1^2+2^2+\cdots+k^2 + (k+1)^2=\frac{1}{6}k(k+1)(2k+1) + (k+1)^2. \] Note that we are trying to prove \(P(k+1)\), so we would like to show that \[1^2+2^2+\cdots+ (k+1)^2=\frac{1}{6}(k+1)(k+1+1)(2(k+1)+1);\nonumber \] we are not quite there yet.
