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Induction: 3+3·5+3·5²+...+3·5^n

Short answer

Base n=0: LHS=3, RHS=3(5−1)/4=3. Assume true for k. For k+1, sum = 3(5^(k+1)−1)/4 + 3·5^(k+1) = 3(5^(k+2)−1)/4. By induction, equality holds for all n≥0.

Use mathematical induction to prove that 3 + 3·5 + 3·5² + ...... + 3·5^n = 3(5^(n+1) − 1)/4 whenever n is a nonnegative integer.

To prove. P(n): 3 + 3·5 + 3·5² + … + 3·5ⁿ = 3(5ⁿ⁺¹ − 1)/4 for every integer n ≥ 0.

Basis step (n = 0)

  • LHS = 3·5⁰ = 3.
  • RHS = 3(5¹ − 1)/4 = 3·4/4 = 3.
  • LHS = RHS, so P(0) is true.

Inductive hypothesis

Assume P(k) is true for some k ≥ 0:

3 + 3·5 + … + 3·5ᵏ = 3(5ᵏ⁺¹ − 1)/4

Inductive step

We must show P(k+1): 3 + 3·5 + … + 3·5ᵏ + 3·5ᵏ⁺¹ = 3(5ᵏ⁺² − 1)/4.

LHS = [3 + 3·5 + … + 3·5ᵏ] + 3·5ᵏ⁺¹
    = 3(5ᵏ⁺¹ − 1)/4 + 3·5ᵏ⁺¹          (inductive hypothesis)
    = 3(5ᵏ⁺¹ − 1)/4 + 12·5ᵏ⁺¹/4       (write 3·5ᵏ⁺¹ over 4)
    = 3(5ᵏ⁺¹ − 1 + 4·5ᵏ⁺¹)/4
    = 3(5·5ᵏ⁺¹ − 1)/4
    = 3(5ᵏ⁺² − 1)/4  = RHS

So P(k) → P(k+1) is true.

Conclusion

By the principle of mathematical induction, P(n) is true for every nonnegative integer n. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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