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Question

C02C1+3C24C3+....+(1)n(n+1)Cn=0.

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Solution

(1+x)n=C0+C1x+C2x2+C3x3+........+Cnxn
(1+x)n×x=C0x+C1x2+C2x3+C3x4+.......+Cnxn+1
Differentisting both sides:
(1+x)n+nx(1+x)n1=C0+2C1x+3C2x2+4C3x3+......+(n+1)Cnxn
Putting x=1
O=C02C1+3C24C3......+(1)x(n+1)CnHence proved

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