Summation of product-pattern series: Find S(n) = 1·2·4 + 2·3·5 + 3·4·6 + ... up to n terms.

Introduction / Context:
Each term of the series is of the form k(k+1)(k+3) for k = 1..n. Converting products into polynomial form allows use of standard summations of k, k^2, and k^3.

Given Data / Assumptions:

• T_k = k(k+1)(k+3) for k ≥ 1.
• We want S(n) = Σ T_k for k = 1..n.

Concept / Approach:
Expand T_k = k(k^2 + 4k + 3) = k^3 + 4k^2 + 3k. Then use standard identities: Σk = n(n+1)/2, Σk^2 = n(n+1)(2n+1)/6, Σk^3 = [n(n+1)/2]^2.

Step-by-Step Solution:
S = Σ(k^3 + 4k^2 + 3k) = Σk^3 + 4Σk^2 + 3Σk= [n^2(n+1)^2]/4 + 4[n(n+1)(2n+1)/6] + 3[n(n+1)/2]= n(n+1)/12 * (3n^2 + 19n + 26)

Verification / Alternative check:
Test small n (e.g., n = 1 or 2) to confirm that both sides match numerically.

Why Other Options Are Wrong:
They miss coefficients from the polynomial expansion or apply an incorrect factor outside the bracket.

Common Pitfalls:
Algebraic slips in expanding k(k+1)(k+3) or forgetting the factor 4 in front of Σk^2.

Final Answer:
[n(n + 1) / 12] (3n2 + 19n + 26)