Question Detail

The expansion of (1 - y)^{-3} = Σ C(n+k-1, k) y^k with n=3. (3 - x)^{-3} = 3^{-3}(1 - x/3)^{-3} = (1/27) Σ C(3+k-1, k) (x/3)^k = (1/27) Σ C(k+2, k) x^k / 3^k. The 3rd term corresponds to k=2 (since first term is k=0). Coefficient = (1/27) × C(4,2) × (1/3²) = (1/27) × 6 × (1/9) = 6/(243) = 2/81. But that's positive. Option C is 2/81, option D is -10/729. The sign? Since the original is (3 - x)^{-3}, the sign alternates? Actually (1 - x/3)^{-3} expansion has all positive terms because it's of the form (1 - y)^{-3} = Σ C(k+2,2) y^k, all positive. So coefficient should be positive 2/81. Option C is 2/81.