Exercise · ZW-MATHEMATICAL-INDUCTION-003 · ★☆☆ Basic

In the inductive step for \(P(n)\), the inductive hypothesis is:

Choose one option.

Show answer

D. \(P(k)\) is true for a fixed \(k\ge n_0\)

Explanation

One assumes \(P(k)\) for an arbitrary but fixed \(k\) and, from it, proves \(P(k+1)\).

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

ZW-MATHEMATICAL-INDUCTION-001 A proof by mathematical induction consists of:
ZW-MATHEMATICAL-INDUCTION-002 To prove 2+4+cdots+2n=n(n+1), in the base case n=1 both sides equal:
ZW-MATHEMATICAL-INDUCTION-004 Using 1+3+cdots+(2n-1)=n^2, what is the sum of the first 12 odd numbers?
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…
ZW-MATHEMATICAL-INDUCTION-009 Let P(n): «n=n+1». It is proved that P(k)⇒ P(k+1) by adding 1 to both sides. What follows?