Question

Why BD=CD

The sum of the series 1^2*3 2^2*3 3^2*4..... to n terms is:
(A) [n^3(n 1)^3 (n 3)]/6
(B) [n(n 1)(3n^2 7n 2)]/12
(C) [n(n 1)]/6*[n(n 1) (2n 1)]
(D) [n(n 1)]/18*[6n(n 1) 2(2n 1)]
(E) None of these

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