The value of ∑0≤i<∑j≤n(nCi+nCj)is:
∑0≤i<∑j≤n(nCi+nCj)
⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩C0+C1+C2+...+Cn +C1+C2+...+Cn +C2+C3+...+Cn ... ... ... +Cn−1+Cn.
=(C0+2C1+3C2+...+nCn−1+n+1Cn−Cn.....(1)
C0x+C1x2+C2x3+....+nCn−1+Cnxn+1=x(1+x)n
Diff. both sides , we get
C0+2C1+3C+...+(n+1)Cnxn
=(1+x)n+xn(1+x)n−1
Putting x = 1 ,
C0+2C1+3C2+....+(n+1)Cn=2n+n2n−1
= (n+2)2n−1
Hence , ∑0≤i<∑j≤n(Ci+Cj) = (n+2)2n−1−1