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Induction: sum of cubes

Short answer

Basis n=1: 1³ = [1·2/2]². Assume sum to k = [k(k+1)/2]². Add (k+1)³, factor to get (k+1)²(k²+4k+4)/4 = [(k+1)(k+2)/2]². Holds by induction.

Use mathematical induction to show that 1³ + 2³ + .... + n³ = [n(n+1)/2]² whenever n is a positive integer.

To prove. P(n): 1³ + 2³ + … + n³ = [n(n+1)/2]² for all positive integers n.

Basis step (n = 1)

  • LHS = 1³ = 1.
  • RHS = [1·2/2]² = 1.
  • P(1) is true.

Inductive hypothesis

Assume P(k) is true: 1³ + 2³ + … + k³ = [k(k+1)/2]² = k²(k+1)²/4.

Inductive step

Show P(k+1): 1³ + … + k³ + (k+1)³ = [(k+1)(k+2)/2]².

LHS = k²(k+1)²/4 + (k+1)³                 (hypothesis)
    = (k+1)² [ k²/4 + (k+1) ]
    = (k+1)² (k² + 4k + 4)/4
    = (k+1)² (k+2)²/4
    = [(k+1)(k+2)/2]²  = RHS

Conclusion

P(1) is true and P(k) → P(k+1), so by mathematical induction P(n) is true for all positive integers n. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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