Suppose that Pascal's triangle is written as follows:
1 1 1 1 1 . . .
1 2 3 4 5 . . .
1 3 6 10 15 . . .
1 4 10 20 35 . . .
1 5 15 35 70 . . .
. . . . .
. . . . .
. . . . .
The first row and column consist entirely of 1s, and every other number is the sum of the number to its left and the number above. For each positive number n, let D(n) denote the determinant of the matrix consisting of the first n rows and first n columns of this array. Compute D(n).
Let L(n) and U(n) be the size n matrices where Pascal's triangle is treated as a lower and upper triangle matrices (n=5 depicted)
[ 1 0 0 0 0 ] [ 1 1 1 1 1 ]
[ 1 1 0 0 0 ] [ 0 1 2 3 4 ]
[ 1 2 1 0 0 ] [ 0 0 1 3 6 ]
[ 1 3 3 1 0 ] [ 0 0 0 1 4 ]
[ 1 4 6 4 1 ] [ 0 0 0 0 1 ]
Then D(n) = L(n)*U(n). Obviously, det(L(n))=1 and det(U(n))=1 for all n.
Then det(D(n)) = det(L(n)*U(n)) = det(L(n))*det(U(n)) = 1*1 = 1 for all n.
See: https://mathworld.wolfram.com/PascalMatrix.html
Quick answer
