Calculate the determinant of the matrix. Calculate the determinant.
\begin{bmatrix} C_{n}^{p+n} & C_{n}^{p+n+1} & \dots & C_{n}^{p+2n} \\ 
C_{n}^{p+n+1} & C_{n}^{p+n+2} & \dots & C_{n}^{p+2n+1} \\ 
\vdots & \vdots & \dots & \vdots \\ 
C_{n}^{p+2n} & C_{n}^{p+2n+1} & \dots & C_{n}^{p+3n} \end{bmatrix}
 A: For $n=0$, we have
$$\det(C(p,0)) = 1$$
For $n=1$, we have
$$\det\left(\begin{bmatrix} p+1 & p+2\\ p+2 & p+3\end{bmatrix} \right) = -1$$
For $n=2$, we have
$$\det\left(\begin{bmatrix} \dfrac{(p+1)(p+2)}2 & \dfrac{(p+2)(p+3)}2 & \dfrac{(p+3)(p+4)}2\\ \dfrac{(p+2)(p+3)}2 & \dfrac{(p+3)(p+4)}2 & \dfrac{(p+4)(p+5)}2\\ \dfrac{(p+3)(p+4)}2 & \dfrac{(p+4)(p+5)}2 & \dfrac{(p+5)(p+6)}2 \end{bmatrix} \right) = -1$$
From a small MATLAB script I wrote, it looks like:
$$D(n,p) = \begin{cases} +1 & \text{if }n \equiv 0,3\pmod4\\ -1 & \text{if } n \equiv 1,2 \pmod4\end{cases}$$
A: Perform the following column and row operations (where A(:,j) and A(i,:) mean resp. the $j$-th column and $i$-th row of $A$):
for k=2 up to k=n
  for j=n down to j=k
    A(:,j) = A(:,j) - A(:,j-1);

for k=2 up to k=n
  for i=n down to i=k
    A(i,:) = A(i,:) - A(i-1,:);

Then we get
$$
A=\begin{bmatrix}
C_{n}^{p+n} & C_{n-1}^{p+n} & \dots & C_{2}^{p+n} & C_{1}^{p+n} & C_{0}^{p+n} \\ 
C_{n-1}^{p+n} & C_{n-2}^{p+n} & \dots & C_{1}^{p+n} & C_{0}^{p+n}\\ 
C_{n-2}^{p+n} & C_{n-3}^{p+n} & \dots & C_{0}^{p+n}\\ 
\vdots & \vdots\\
C_{1}^{p+n} & C_{0}^{p+n}\\
C_{0}^{p+n}
\end{bmatrix},
$$
where the strictly lower anti-triangular part of $B$ is zero and all antidiagonal entries are equal to $1$. Hence $\det(A)$ is equal to the determinant of the mirror image of $I_{n+1}$, which is $(-1)^{n(n+1)/2}$. (Note that $A$ is $(n+1)\times(n+1)$, not $n\times n$).
