Exercise · ZW-MATHEMATICAL-INDUCTION-005 · ★☆☆ Basic
In the inductive step of \(1+2+\cdots+n=\tfrac{n(n+1)}{2}\), adding \(k+1\) to both sides of \(P(k)\) makes the right side equal to:
Choose one option.
Show answer
- A. \(\tfrac{(k+1)(k+2)}{2}\)
Explanation
We have \(\tfrac{k(k+1)}{2}+(k+1)=(k+1)\big(\tfrac k2+1\big)=\tfrac{(k+1)(k+2)}{2}\), which is \(P(k+1)\). Adding just \(1\) forgets that the new term is \(k+1\).
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-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-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?
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