Compute \[ \Delta_n= \det\left[\binom{i+j}{i}\right]_{0\le i,j\le n-1}. \]
Solution
Let \[ A_n \coloneqq \left[\binom{i+j}{i}\right]_{0\leq i,j\leq n-1}, \] so that \(\Delta_n=\det A_n\).
1. Starting with \(A_n\), for \(j=n-1,n-2,\ldots,1\) replace column \(j\) by\[ C_j-C_{j-1}. \]The resulting matrix \(B_n=(b_{ij})\) has\[ b_{i0}=1, \]and, for \(j\geq1\),\[ b_{0j}=0, \qquad b_{ij}=\binom{i+j-1}{i-1} \quad(i\geq1). \]
Each column replacement preserves the determinant. The columns are changed in descending order, so when column \(j\) is replaced, column \(j-1\) still has its original entries. Thus, for \(j\geq1\) and \(i\geq1\), \[ \begin{aligned} b_{ij} &= \binom{i+j}{i} - \binom{i+j-1}{i}\\ &= \binom{i+j-1}{i-1} \end{aligned} \] by Pascal’s identity. For \(i=0\) the same column difference is \[ 1-1=0. \] Column \(0\) is unchanged and consists entirely of \(1\)’s.
2. Starting with \(B_n\), for \(i=n-1,n-2,\ldots,1\) replace row \(i\) by\[ R_i-R_{i-1}. \]The resulting matrix is\[ \begin{pmatrix} 1&0\\ 0&A_{n-1} \end{pmatrix}. \]
Again, each row replacement preserves the determinant, and descending order ensures that row \(i-1\) is still unchanged when it is subtracted from row \(i\).
By step 1, the first row of \(B_n\) is \((1,0,\ldots,0)\) and the first column consists of \(1\)’s. Hence after the row differences, the first column becomes \((1,0,\ldots,0)^T\) while the first row stays fixed.
For \(i\geq2\) and \(j\geq1\), the new \((i,j)\) entry is \[ \begin{aligned} \binom{i+j-1}{i-1} - \binom{i+j-2}{i-2} &= \binom{i+j-2}{i-1}, \end{aligned} \] by Pascal’s identity. For \(i=1\) and \(j\geq1\), the new entry is \[ \binom{j}{0}-0 = 1 = \binom{j-1}{0}, \] so the same final formula holds for every \(i,j\geq1\). Reindexing by \[ i'=i-1, \qquad j'=j-1 \] shows that the lower-right block has entries \[ \binom{i'+j'}{i'}, \] which is exactly \(A_{n-1}\).
3. For every \(n\geq2\),\[ \Delta_n=\Delta_{n-1}. \]
The column and row operations in steps 1 and 2 preserve determinant. Therefore \[ \Delta_n = \det \begin{pmatrix} 1&0\\ 0&A_{n-1} \end{pmatrix} = \det A_{n-1} = \Delta_{n-1}. \]
4. One has\[ \Delta_1=1. \]
The matrix \(A_1\) is the \(1\times1\) matrix \[ \left[\binom{0}{0}\right]=[1]. \]
5. Therefore, for every \(n\geq1\),\[ \boxed{\Delta_n=1}. \]
Apply step 3 repeatedly until reaching the base case in step 4.
6. Q.E.D.
Step 5 gives the requested determinant.