The binomial series
Problem 7.596 · medium
Find the first five terms of the binomial series for \( \displaystyle \frac{1}{\sqrt[3]{\frac{x}{2} + 1}} \).
- (1 + u)ʳ = 1 + ru + r(r − 1)/2! u² + r(r − 1)(r − 2)/3! u³ + ⋯ with r = -1/3 and u = x/2.
- \[ \frac{\left(1 \cdot \frac{1}{2}\right)^{2} \left(1 \cdot \frac{1}{3} + 1\right)}{2 \cdot 3} = \frac{1}{18} \]The x² coefficient, r(r − 1)/2 · c².✓ Proved
- \[ \lim_{x \to 0}\left(\frac{- \frac{35 x^{4}}{3888} + \frac{7 x^{3}}{324} - \frac{x^{2}}{18} + \frac{x}{6} - 1 + \frac{1}{\sqrt[3]{\frac{x}{2} + 1}}}{x^{4}}\right) = 0 \]These terms agree with the function to order x⁴.✓ Proved
Answer \( \frac{35 x^{4}}{3888} - \frac{7 x^{3}}{324} + \frac{x^{2}}{18} - \frac{x}{6} + 1 + \cdots \)
Lines: 2 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt. How we verify
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | f minus the polynomial at x = 0.01 and 0.001 |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The stated answer has incorrect signs for the linear and cubic terms. The binomial expansion of (1 + x/2)^{-1/3} yields -x/6 for the linear term and -7x^3/324 for the cubic term, but the stated answer lists them as positive. The solution's verification step (line 3) checks the negative of the stated answer against the function, implying the stated answer is wrong, yet it claims the terms agree.
Every verdict on record (6)
qwen3.6:27b-mlx: fail (error) 2026-10-17 — The stated answer has incorrect signs for the linear and cubic terms. The binomial expansion of (1 + x/2)^{-1/3} yields -x/6 for the linear term and -7x^3/324 for the cubic term, but the stated answer lists them as positive. The solution's verification step (line 3) checks the negative of the stated answer against the function, implying the stated answer is wrong, yet it claims the terms agree.gpt-oss:20b: pass 2026-10-17qwen3.6:27b-mlx: fail (error) 2026-10-15 — The stated answer has incorrect signs for all non-constant terms. The binomial expansion of (1 + x/2)^(-1/3) yields alternating signs starting with a negative linear term (-x/6), but the stated answer lists -x/6 + x^2/18 - 7x^3/324 + 35x^4/3888, which is actually the expansion for (1 - x/2)^(-1/3) or simply has the signs flipped relative to the standard expansion where u=x/2. Wait, let's re-evaluate. r = -1/3. Term 1: 1. Term 2: (-1/3)(x/2) = -x/6. Term 3: (-1/3)(-4/3)/2 * (x/2)^2 = (4/9)/2 * x^2/4 = (2/9)*(x^2/4) = x^2/18. Term 4: (-1/3)(-4/3)(-5/3)/6 * (x/2)^3 = (-20/27)/6 * x^3/8 = (-20/162) * x^3/8 = (-10/81) * x^3/8 = -10x^3/648 = -5x^3/324. The stated answer has -7x^3/324. Term 5: (-1/3)(-4/3)(-5/3)(-6/3)/24 * (x/2)^4 = (120/81)/24 * x^4/16 = (5/27) * x^4/16 = 5x^4/432. The stated answer has 35x^4/3888. 5/432 = 35/3024. 35/3888 is different. The coefficients in the stated answer are incorrect. Specifically, the coefficient for x^3 is -5/324, not -7/324. The coefficient for x^4 is 5/432, not 35/3888. The solution claims the terms agree to order x^4, but the coefficients provided in the 'Stated answer' are wrong.gpt-oss:20b: pass 2026-10-15qwen3.6:27b-mlx: inconclusive 2026-10-14 — reviewer response could not be parsed: { "verdict": "fail", "severity": "error", "notes": "The stated answer has incorrect signs for the linear and cubic terms. The binomial expansion of (1 + x/2)^(-1/3) yields -x/6 for the linear term, bugpt-oss:20b: pass 2026-10-14
Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate
method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence,
not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by
generator:structured/binomial_series, checked 2026-10-17 with SymPy 1.14.0.