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Question

Suppose det ⎢ ⎢ ⎢ ⎢nk=0knk=0nCk k2nk=0nCk knk=0nCk 3k⎥ ⎥ ⎥ ⎥=0
hold for some positive integer n. Then nk=0nCkk+1 equals

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Solution

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)2n2n.2n14n]=0

n(n+1)2n1∣ ∣122n2n2n+1∣ ∣=0

n(n+1)22n3∣ ∣121n8∣ ∣=0
n(n+1)22n3(4n)=0
Hence, n=0,1,4n=4 [n>0]
4k=04Ckk+1=4C01+4C12+4C23+4C34+4C45=315=6.2

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