Mathematical Induction Math Example 4

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Example 4

medium
Prove by induction that 2+4+6+⋯+2n=n(n+1)2 + 4 + 6 + \cdots + 2n = n(n+1) for all n≥1n \ge 1.

Solution

  1. 1
    Base case n=1n=1: the left side is 2 and the right side is 1(1+1)=21(1+1)=2, so the statement is true.
  2. 2
    Assume true for n=kn=k: 2+4+⋯+2k=k(k+1)2+4+\cdots+2k = k(k+1). For n=k+1n=k+1, add 2(k+1)2(k+1) to both sides: k(k+1)+2(k+1)=(k+1)(k+2)k(k+1)+2(k+1) = (k+1)(k+2), which matches (k+1)((k+1)+1)(k+1)((k+1)+1).

Answer

2+4+6+⋯+2n=n(n+1)for all n≥12 + 4 + 6 + \cdots + 2n = n(n+1) \quad \text{for all } n \ge 1
Induction proves the first case and then shows the formula survives one more step. Once both parts are established, the result follows for every positive integer.

About Mathematical Induction

Mathematical induction proves statements indexed by integers by verifying a base case and an inductive step.

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More Mathematical Induction Examples

Example 1 medium

Use mathematical induction to prove: [formula] for all [formula].

Example 2 hard

Prove by induction: [formula] for all integers [formula].

Example 3 medium

Prove by induction: [formula] for all [formula].

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