Let `a_k = .^nC_k` for `0 lt= k lt= n` and `A_k = [(a_(k-1),0), (0,a_k)]` for `1 lt= k lt= n` and `B=sum_(k=1)^(n-1)A_k * A_(k+1) = [(a,0), (0,b)]`

Answer

1

Explanation

`A_(k).A_(k+1)=[(a_(k-1), 0),(0,a_(k))][(a_(k),0),(0,a_(k+1))]`
`A_(k).A_(k+1)=[(a_(k-1).a_(k), 0),(0,a_(k)a_(k+1))]`
`A_(k).A_(k+1)=[(.^(n)C_(k-1) .^(n)C_(k), 0),(0, .^(n)C_(k) .^(n)C_(k+1))]`
`B=[(sum_(k=1)^(n=1).^(n)C_(k-1).^(n)C_(k),0),(0,sum_(k=1)^(n-1).^(n)C_(k).^(n)C_(k+1))]=[(a,0),(0,b)]`
`a=b=.^(2n)C_(n-1-n)`