The correct option is A n2⋅2n
We have,
∑∑0≤r,s≤n(r+s)(Cr+Cs)
=n∑r=0n∑s=0(rCr+rCs+sCr+sCs)
=n∑r=0[n∑s=0rCr+n∑s=0rCs+n∑s=0sCr+n∑s=0sCs]
=n∑r=0[(n+1)rCr+r2n+n(n+1)2Cr+n⋅2n−1]
=(n+1)(n⋅2n−1)+2nn(n+1)2+n(n+1)22n+n⋅2n−1(n+1)
=n(n+1)2n+n(n+1)2n
=2n(n+1)2n
Also
∑∑0≤r,s≤n(r+s)(Cr+Cs)
=n∑r=04rCr+2∑∑0≤r<s≤n(r+s)(Cr+Cs)
∴2n(n+1)2n=4n⋅2n−1+2∑∑0≤r<s≤n(r+s)(Cr+Cs)
⇒∑∑0≤r<s≤n(r+s)(Cr+Cs)=n2⋅2n