Binomial Theorem 2 Question 14

17. Prove that C022C1+32C2+(1)n(n+1)2Cn =0,n>2, where Cr=nCr.

(1989,5M)

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Solution:

  1. C022C1+32C2+(1)n(n+1)2Cn

=r=0n(1)r(r+1)2nCr=r=0n(1)r(r2+2r+1)nCr

=r=0n(1)rr2nCr+2r=0n(1)rrnCr+r=0n(1)rnCr=r=0n(1)rr(r1)nCr+3r=0n(1)rrnCrn

+r=0n(1)rnCr

=r=2n(1)rn(n1)n2Cr2+3r=1n(1)rnn1Cr1 +r=0n(1)rnCr

=n(n1)n2C0n2C1+n2C2+(1)nn2Cn2

+3nn1C0+n1C1n1C2++(1)nn1Cn1

+nC0nC1+nC2++(1)nnCn

=n(n1)0+3n0+0,n>2=0,n>2



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