lean_workbook_plus_16619

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Proved status carries no information about that problem. Do not import this node or use it as a dependency.

since, $a_{n+1}^3=99a_3^3$ for every positive integer $n \geq 1.$ We have,

Therefore, $a_n= 99^{\frac{n-1}{3}}$ for every positive integer $n \geq 1.$ Thus, $a_{100}=99^\frac{100-1}{3}=99^{33}.$ Which corresponds to answer choice $\textbf{(C)}$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_16619  (a : ℕ → ℝ)
  (h₀ : 0 < a 1)
  (h₁ : ∀ n, a (n + 1)^3 = 99 * (a 1)^3)
  (h₂ : ∀ n, 0 < a n) :
  a 100 = 99^33   :=  by sorry

The recurrence uses the fixed a(1) and includes n=0, instead of the preceding term in the source progression.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

Use a(1)=1 and a(n+1)^3=99*a(n)^3 only for n≥1, with positive terms, then conclude a(100)=99^33.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.