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Induction: alternating sum of squares

Short answer

Basis n=1: 1² = 1 = (−1)²·1·2/2. Assume sum to k equals (−1)^(k+1)k(k+1)/2. Adding (−1)^(k+2)(k+1)² and factoring yields (−1)^(k+2)(k+1)(k+2)/2. By induction, the formula holds.

Use mathematical induction to verify: 1² − 2² + 3² − 4² + ... + (−1)^(n+1) n² = (−1)^(n+1) n(n+1)/2

To prove. P(n): 1² − 2² + 3² − 4² + … + (−1)ⁿ⁺¹ n² = (−1)ⁿ⁺¹ n(n+1)/2 for all positive integers n.

Basis step (n = 1)

  • LHS = 1² = 1.
  • RHS = (−1)² · 1 · 2 / 2 = 1.
  • So P(1) is true.

Inductive hypothesis

Assume P(k) is true for some k ≥ 1:

1² − 2² + … + (−1)ᵏ⁺¹ k² = (−1)ᵏ⁺¹ k(k+1)/2

Inductive step

We show P(k+1): the sum up to (−1)ᵏ⁺² (k+1)² equals (−1)ᵏ⁺² (k+1)(k+2)/2.

LHS = [1² − 2² + … + (−1)ᵏ⁺¹ k²] + (−1)ᵏ⁺² (k+1)²
    = (−1)ᵏ⁺¹ k(k+1)/2 + (−1)ᵏ⁺² (k+1)²     (hypothesis)

Since (−1)ᵏ⁺¹ = −(−1)ᵏ⁺², take (−1)ᵏ⁺² (k+1) as common factor:

    = (−1)ᵏ⁺² (k+1) [ −k/2 + (k+1) ]
    = (−1)ᵏ⁺² (k+1) (k+2)/2
    = RHS of P(k+1)

Conclusion

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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