∫Calc Practice

Sums with sigma notation

Problem 4.1157 · medium

Evaluate \( \displaystyle \sum_{i=3}^{15} \left(2 i^{2} - 4 i\right) \) using summation formulas.
  1. Split the sum and use Σᵢ₌₁ᴺ 1 = N, Σ i = N(N + 1)/2, Σ i² = N(N + 1)(2N + 1)/6, Σ i³ = [N(N + 1)/2]².
  2. The sum starts at i = 3: compute from 1 to 15 and subtract the sum from 1 to 2.
  3. \[ \left(-4\right) 15 \cdot 16 \cdot 1 \cdot \frac{1}{2} + 2 \cdot 15 \cdot 16 \cdot 31 \cdot 1 \cdot \frac{1}{6} = 2000 \]
    From i = 1 to 15.✓ Proved
  4. \[ \sum_{i=1}^{2} \left(2 i^{2} - 4 i\right) = -2 \]
    From i = 1 to 2.✓ Proved
  5. \[ 2 + 2000 = 2002 \]
    Subtract.✓ Proved
  6. \[ \sum_{i=3}^{15} \left(2 i^{2} - 4 i\right) = 2002 \]
    The sum, evaluated directly, agrees.✓ Proved
Answer \( 2002 \)

Lines: 4 proved, 2 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
LineStatusChecked byDetail
1Not checked—a sentence; read, not computed
2Not checked—a sentence; read, not computed
3✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
4✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
5✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
6✓ Provedsympy 1.14.0simplify(lhs - rhs) reduced to 0
answer, a second way✓ Checked independentlysympy 1.14.0 + mpmath 1.3.0the terms added one by one in Python integers

Reviewers

  • gpt-oss:20b: fail (misleading) — Sentence 5 incorrectly states "2 + 2000 = 2002" while the correct computation is 2000 - (-2) = 2002. This could mislead a student into thinking addition is used instead of subtraction.
  • qwen3.6:27b-mlx: fail (error) — Step 5 claims to subtract the sum from 1 to 2, but performs addition (2000 + 2) instead of subtraction (2000 - (-2)). While the arithmetic result 2002 is coincidentally correct because -(-2) = +2, the sentence describing the operation is mathematically false and misleading.
Every verdict on record (6)
  • qwen3.6:27b-mlx: fail (error) 2026-10-18 — Step 5 claims to subtract the sum from 1 to 2, but performs addition (2000 + 2) instead of subtraction (2000 - (-2)). While the arithmetic result 2002 is coincidentally correct because -(-2) = +2, the sentence describing the operation is mathematically false and misleading.
  • gpt-oss:20b: fail (misleading) 2026-10-18 — Sentence 5 incorrectly states "2 + 2000 = 2002" while the correct computation is 2000 - (-2) = 2002. This could mislead a student into thinking addition is used instead of subtraction.
  • qwen3.6:27b-mlx: fail (error) 2026-10-16 — The solution claims to subtract the sum from 1 to 2, but Step 5 adds 2 instead of subtracting -2 (i.e., it computes 2000 + 2 instead of 2000 - (-2)). While the final numerical answer happens to be correct due to a sign error in the subtraction logic cancelling out the negative value of the partial sum, the reasoning 'Subtract' leading to '2 + 2000' is mathematically incorrect and misleading.
  • gpt-oss:20b: pass 2026-10-16
  • qwen3.6:27b-mlx: fail (error) 2026-10-15 — Step 5 contains an arithmetic error: subtracting the sum from 1 to 2 (-2) from the sum from 1 to 15 (2000) requires calculating 2000 - (-2) = 2002, but the step writes '2 + 2000', implying the subtracted sum was +2. While the final number is correct by coincidence (since -(-2) = +2), the logic presented in the equation is flawed because it treats the value to be subtracted as positive 2 rather than negative 2, or confuses the operation. Specifically, the text says 'Subtract' but performs addition of a positive 2. A clearer derivation would be 2000 - (-2) = 2002.
  • gpt-oss:20b: pass 2026-10-15

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/summation_formulas, checked 2026-10-18 with SymPy 1.14.0.