Exercise · ZW-MATHEMATICAL-INDUCTION-010 · ★★★ Advanced
\(P(3)\) is checked and \(P(k)\Rightarrow P(k+1)\) is proved for all \(k\ge3\). For which \(n\) is \(P(n)\) proved?
Choose one option.
Show answer
- B. For all \(n\ge3\)
Explanation
The chain starts at the base case \(3\) and moves on to \(4,5,\dots\); it says nothing about \(n=1,2\).
Review the concept: Mathematical induction
Mathematical induction proves that a property \(P(n)\) holds for every integer \(n\ge n_0\): one checks the base case \(P(n_0)\) and proves that, for each \(k\ge n_0\), \(P(k)\) implies \(P(k+1)\).
\[\big[P(n_0)\wedge\forall k\ge n_0\,(P(k)\Rightarrow P(k+1))\big]\Rightarrow\forall n\ge n_0\,P(n)\]
More practice problems: mathematical induction
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- ZW-MATHEMATICAL-INDUCTION-002 To prove 2+4+cdots+2n=n(n+1), in the base case n=1 both sides equal:
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- ZW-MATHEMATICAL-INDUCTION-005 In the inductive step of 1+2+cdots+n=tfracn(n+1)2, adding k+1 to both sides of P(k) makes the right…
- ZW-MATHEMATICAL-INDUCTION-006 Using 1^2+2^2+cdots+n^2=tfracn(n+1)(2n+1)6, compute 1^2+2^2+cdots+6^2.
- ZW-MATHEMATICAL-INDUCTION-007 To prove that 6 divides 7^n-1, which equality lets you go from P(k) to P(k+1)?
- ZW-MATHEMATICAL-INDUCTION-008 The inequality 2^n>n^2 fails for n=2,3,4 and holds for n≥5. Which base case is suitable to prove it…