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Question

Prove that 1nC11+x1+nx+nC2 1+2x1+nx2nC3 1+3x1+nx3+.....(n+1) terms = 0

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Solution

Put 11+nx=t and split into two .
(C0C1t+C2t2C3t3+...)+x(C1t+2C2t23c3t3+....)=A+xB
A=C0C1t+C2t2+C3t3+.... ......(1)
(1t)n=(111+nx)n=(nx1+nx)n
In order to get B , we differentiate (1)
n(1t)n1=C1+2C2t3C3t2+....
Multiply both sides by t
n(1t)n1=C1+2C2t3C3t2+....=B
B=n11+nx[111+nx]n1
=n1+nx(nx1+nx)n1 ....(2)
A+xB
=(nx1+nx)nnx1+nx(nx1+nx)n1
=(nx1+nx)n(nx1+nx)n=0

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