Exercise · ZW-MATHEMATICAL-INDUCTION-008 · ★★☆ Intermediate
The inequality \(2^n>n^2\) fails for \(n=2,3,4\) and holds for \(n\ge5\). Which base case is suitable to prove it by induction?
Choose one option.
Show answer
- D. \(n=5\)
Explanation
The base case must be the first \(n\) from which the property holds: \(n_0=5\), since \(32>25\). The cases \(n=2,3,4\) violate it.
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-003 In the inductive step for P(n), the inductive hypothesis is:
- 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-009 Let P(n): «n=n+1». It is proved that P(k)⇒ P(k+1) by adding 1 to both sides. What follows?
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