Hiiiiiiiiiiiiiii again 
...here's the solution
(1+x)n = C0 + C1x + C2 x2 +............
Differentiating once wrt x
n(1+x)n-1 = C1 + 2C2 x +............ ....eqn1
Multipyling both the sides of equation by x and then differentiating once again wrt x, we have
n[(1+x)n-1 + x(n-1) (1+x)n-2] =12 C1 + 22 C2 x +............ + n2 Cn xn-1 .... eqn2
reversing the eqn 1's RHS, n then taking the coeffecient of xn-1 in the expression after multiplying with eqn 2, we get
12 n C1 Cn + 22 (n-1) C2 C n-1 +.......... 2(n-1)C2 C n-1 + 1.n2 C1 C n
= coeffecient of xn-1 in n(1+x)n-1 {n[(1+x)n-1 + x(n-1) (1+x)n-2]}
= [n2 (n+1)(2n-3)!] / [(n-1)! (n-2)!]
hopefully this is the answer, if m right do rate me!!!
Moderation Team
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