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Induction: sum of geometric progression

Short answer

Basis n=0: a = (ar−a)/(r−1). Assume sum to k is (ar^(k+1)−a)/(r−1). Adding ar^(k+1) gives (ar^(k+2)−a)/(r−1). By induction, true for all n≥0.

Use mathematical induction to prove the formula for the sum of a finite number of terms of Geometric Progression: Σ(j=0 to n) a·r^j = a + ar + ar² + .... + ar^k = (a·r^(n+1) − a)/(r − 1), when r ≠ 1, where n is non-negative integer.

(The last term written as ar^k in the question should be arⁿ, since the sum runs from j = 0 to n.)

To prove. P(n): a + ar + ar² + … + arⁿ = (arⁿ⁺¹ − a)/(r − 1), for r ≠ 1 and all integers n ≥ 0.

Basis step (n = 0)

  • LHS = ar⁰ = a.
  • RHS = (ar − a)/(r − 1) = a(r − 1)/(r − 1) = a, since r ≠ 1.
  • P(0) is true.

Inductive hypothesis

Assume P(k) is true: a + ar + … + arᵏ = (arᵏ⁺¹ − a)/(r − 1).

Inductive step

Show P(k+1): a + ar + … + arᵏ + arᵏ⁺¹ = (arᵏ⁺² − a)/(r − 1).

LHS = (arᵏ⁺¹ − a)/(r − 1) + arᵏ⁺¹            (hypothesis)
    = [arᵏ⁺¹ − a + arᵏ⁺¹ (r − 1)]/(r − 1)
    = [arᵏ⁺¹ − a + arᵏ⁺² − arᵏ⁺¹]/(r − 1)
    = (arᵏ⁺² − a)/(r − 1) = RHS

Conclusion

By mathematical induction, the formula holds for every non-negative integer n. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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