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Q.

knC1(k1)+nC2(k2)nC3(k3)++(1)nnCn(kn)=  

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An Intiative by Sri Chaitanya

a

0

b

1

c

2

d

3

answer is D.

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Detailed Solution

knC1(k1)+nC2(k2)nC3(k3)++(1)nnCn(kn)

=knC1k+nC1+nC2.k2.nC2nC3.k+3nC3++(1)n.nCn.k(1)n.nCn.n

=k[nC0nC1+nC2nC3++(1)n.nCn]+nC12nC2+3nC3++(1)n.n.nCn

=k(0)+0=0

(1+x)n=C0+C1x+C2x2++Cnxn

put x=1

=C0C1+C2C3++(1)n.Cn

Differentiative

n(1+x)n1=C1+2C2.x+3C3x2+nCnxn1

put x=1

0=C12C2+3.C3++(1)nnCn

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