det [0+1+2+...+nnC0(02)+nC1(12)+nC2(22)+...+nCn(n2)nC0(0)+nC11+nC22+...+nCnnnC0(30)+nC1(31)+nC2(32)+...+nCn(3n)]=0
⇒det [n(n+1)2n(n+1)2n−2n.2n−14n]=0
⇒n(n+1)2n−1∣∣
∣∣122n−2n2n+1∣∣
∣∣=0
⇒n(n+1)22n−3∣∣
∣∣121n8∣∣
∣∣=0
⇒n(n+1)22n−3(4−n)=0
Hence, n=0,−1,4⇒n=4 [∵n>0]
∴4∑k=04Ckk+1=4C01+4C12+4C23+4C34+4C45=315=6.2