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Question

The value of 0i<jn(nCi+nCj)is: 



Solution

The correct option is D


0i<jn(nCi+nCj) 

⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪C0+C1+C2+...+Cn     +C1+C2+...+Cn               +C2+C3+...+Cn                  ...       ...     ...                             +Cn1+Cn.

=(C0+2C1+3C2+...+nCn1+n+1CnCn.....(1) 

      C0x+C1x2+C2x3+....+nCn1+Cnxn+1=x(1+x)n 

Diff. both sides , we get 

C0+2C1+3C+...+(n+1)Cnxn 

                                          =(1+x)n+xn(1+x)n1

Putting x = 1 , 

C0+2C1+3C2+....+(n+1)Cn=2n+n2n1 

                                                                       = (n+2)2n1

Hence , 0i<jn(Ci+Cj)  = (n+2)2n11 

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