$${ a }_{ 1,n }={ a }_{ n,1 }=1\quad \forall \quad n $$ $${ a }_{ i+1,j+1 }={ a }_{ i,j }+{ a }_{ i,j+1 }+{ a }_{ i+1,j }$$
Let $A$ be the matrix ($k$ order) made by using ${ a }_{ i,j }$ as element in $i$th row and $j$th column,in other words $A={ \left[ { a }_{ i,j } \right] }_{ k*k }$
Find the determinant of matrix $A$ in terms of $k$
For example, for $k=4$, $A=\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 3 & 5 & 7 \\ 1 & 5 & 13 & 25 \\ 1 & 7 & 25 & 63 \end{bmatrix}$
I calculated some determinants and found that the determinants form a series $2^0,2^1,2^3,2^6,{2}^{10}...$
Looking at the patterns the determinant should be ${2}^{(k-1)(k)/2}$ but i have no way of doing that
According to How to proceed with this multi-variable recurrence? and A008288 in the OEIS, the $a_{ij}$ are the Delannoy numbers (only indexed differently), and we have the formula $$ a_{i,j} = D(i-1, j-1) = \sum_{d=1}^k \underbrace{\binom{i-1}{d-1}}_{b_{i,d}} \cdot 2^d \cdot \underbrace{\binom{j-1}{d-1}}_{b_{j, d}} \, . $$ It follows that $A = B D B^T$ where $B$ is a lower triangular matrix (of binomial coefficients) with $(1, \ldots, 1)$ on the diagonal, and $D$ is a diagonal matrix with $(2^0,\ldots,2^{k-1})$ on the diagonal. Therefore $$ \det(A) = \det(D) = 2^0 \cdot 2^1 \cdots 2^{k-1} = 2^{(k-1)k/2} $$ as you conjectured.
As an example, for $k=4$: $$ A=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 \\ 1 & 2 & 1 & 0 \\ 1 & 3& 3 & 1 \end{bmatrix} \times \begin{bmatrix} 2^0 & 0 & 0 & 0 \\ 0 & 2^1 & 0 & 0 \\ 0 & 0 & 2^2 & 0 \\ 0 & 0& 0 & 2^3 \end{bmatrix} \times \begin{bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 1 & 3 \\ 0 & 0 & 0 & 1 \end{bmatrix} $$