Select the correct prismoidal (Simpson’s) formula for volume between a series of equally spaced cross-sections: identify the proper weighting of odd and even intermediate areas relative to the first and last areas.

Introduction / Context:
In earthwork/highway/river training projects, volume between a sequence of equally spaced cross-sections is frequently computed using the prismoidal (Simpson’s) formula. Mastering the correct weighting of odd and even intermediate areas avoids systematic over- or underestimation compared to the trapezoidal rule.

Given Data / Assumptions:

• Sections are at equal spacing D.
• Areas: A0 (first), An (last), with intermediate A1, A2, ..., An−1.
• We seek the correct coefficient pattern for odd and even indices.

Concept / Approach:
Simpson’s (prismoidal) rule for equally spaced sections is V = (D/3) * [A0 + An + 4 * (sum of odd-index areas) + 2 * (sum of even-index areas)]. The hallmark of Simpson’s rule is the 4–2 alternating weights starting with 4 on A1 (odd) and ending with 4 on A_{n−1} (also odd when n is even).

Step-by-Step Solution:
1) Write the canonical form: V = (D / 3) * [A0 + An + 4 Σ A_odd + 2 Σ A_even].2) Identify that odd-index areas (A1, A3, ...) receive weight 4.3) Even-index areas (A2, A4, ...) receive weight 2.4) Among the options, only the expression with '2 ∑ Even' and '4 ∑ odd' has the correct relative weighting pattern.

Verification / Alternative check:
Comparing against the trapezoidal rule V = (D/2) [A0 + An + 2 Σ A_i] confirms Simpson’s extra curvature correction via alternating 4–2 weights, which reduces error for parabolic area variation.

Why Other Options Are Wrong:

• Option A mis-weights even/odd and lacks the /3 factor notionally implied.
• Option B gives 4 to even areas, which is reversed.
• Option D is the same pattern but punctuation aside; the core correct pattern is '2 even, 4 odd' (captured by Option C).

Common Pitfalls:

• Swapping the 4 and 2 multipliers.
• Applying Simpson’s rule to an odd number of intervals without adjustment.

Final Answer:
[first area + last area + 2 ∑ Even area + 4 ∑ odd areas].