Suppose det∑k=0nk        ∑k=0nCk   nk2∑k=0nnCkk∑k=0nnCk3k=0 Holds for some positive integer n. The ∑k=0n nCkk+1 equals

n(n+1)2n(n-1)2n-2+n2n-1n2n-14n=0⇒n(n+1)22n-1-n(n-1)22n-3-n22n-2=0Cancel out 22n-3 from each term as common term ⇒4(n+1)-n(n-1)-2n=0⇒n2-3n-4=0⇒n=4(∵n>0,n∈N)∑k=04Ck   4k+1=25-15=315=6.20