Help getting a sum into another form I have the following sum:
$$\sum_{i=0}^n\sum_{j=0}^{n-i}a_{i,j}r^is^jt^{n-i-j}u^{2n-2i-j},$$
which I'd like to get into the form
$$\sum_{k=0}^{2n}\lambda_ku^k.$$
In the case $n=3$ I have formed the following analysis of the indices table:
$$\begin{array}{c|c|c}
i&j&6-2i-j\\
\hline
0&0&6\\
0&1&5\\
0&2&\textbf{4}\\
0&3&\textbf{3}\\
\hline
1&0&\textbf{4}\\
1&1&\textbf{3}\\
1&2&\textit{2}\\
\hline
2&0&\textit{2}\\
2&1&1\\
\hline
3&0&0
\end{array}$$
The bold and italic terms could clearly be grouped together, so obviously yes it can be written in the form described, but is there a nice formula for any $n$ ?
 A: Given S.Koch's comment above I was able to do the following, which I should've been able to do anyway as I used to use these sums quite a bit but a long time ago. We have:
$$\sum_{i=0}^n\sum_{j=0}^{n-i}a_{i,j}r^is^jt^{n-i-j}u^{2n-2i-j},$$
changing the indexing in the inner sum slightly gives
$$\sum_{i=0}^n\sum_{j=i}^na_{i,j-i}r^is^{j-i}t^{n-j}u^{2n-(i+j)}.$$
Now, taking a look at the index set for $n=3$ we see the following:
$$\begin{array}{l}i\\\hline0,1,2,3,4,5,6\\1,2,3,4,5,6\\2,3,4,5,6\\3,4,5,6\\4,5,6\\5,6\\6\end{array},$$
with $j$ running down the rows. We can traverse these instead over the diagonals such that $i+j=k$ with $k=0,1,2,3,4,5,6$, hence
$$\sum_{k=0}^{2n}\left(\sum_{i+j=k}a_{i,j-i}r^is^{j-i}t^{n-j}\right)u^{2n-k}.$$
Expanding the index of the inner sum, $$\sum_{k=0}^{2n}\left(\sum_{i=0}^ka_{i,k-2i}r^is^{k-2i}t^{n-k+i}\right)u^{2n-k},$$
then substituting $k$ with $2n-k$ finally gives
$$\sum_{k=0}^{2n}\left(\sum_{i=0}^{2n-k}a_{i,2n-k-2i}r^is^{2n-k-2i}t^{i+k-n}\right)u^k,$$
as required.
