I'm trying to compute the double sum :
$$ \sum_{1\leq i, j\leq n, \ i+j\leq n }\binom{i+j}{i} x^i y^j $$
Here $(x,y) \in \mathbb{R}^2$, although it is not mentioned in the source.
Note: a typo in the original question was to evaluate the sum for $1 \le i \le j \le n$ instead of $1 \le i,j \le n$.
I've thought about Newton's theorem, about rewriting it as : $$ \sum_{j=1}^{n} \sum_{i=1, i+j \leq n}^j \begin{pmatrix}i+j\\ j \end{pmatrix} x^iy^j $$
But it was in vain, the derivatives either do not lead to something helpful.
I know how to answer the question if the $i\leq j$ condition is dropped (perhaps that’s a typo in your question ?). If one denotes the sum by $S$, one can write
$$ \begin{array}{lcl} S&=&\sum_{t=2}^{n} \sum_{1\leq i,j\leq n, i+j=t }\binom{i+j}{i}x^iy^{j}= \sum_{t=2}^{n} \sum_{i=1}^{t-1}\binom{t}{i}x^iy^{t-i} \\ &=& \sum_{t=2}^{n} (x+y)^t-(x^t)-(y^t)=\frac{(x+y)^2-(x+y)^{n+1}}{1-(x+y)}- \frac{x^2-x^{n+1}}{1-x}-\frac{y^2-y^{n+1}}{1-y} \end{array} $$
Hope this helps.