I'm trying to multiply two different finite summations with different upper limits. I've tried Cauchy Product but i think it's valid for same upper limits. I also tried to split the summation. Any ideas?
Edit: I'm trying to multiply $$ \bigg[ \sum_{k=0}^{\alpha +1} \binom{\alpha+1}{k} \cdot x{_{i}}^{\alpha-k+1} \cdot (x_{i+1}-x_{i})^k \cdot t^k \bigg] \cdot \bigg[\sum_{j=0}^{\beta} \binom{\beta}{j} \cdot y{_{i}}^{\beta-j} \cdot (y_{i+1}-y_{i})^j \cdot t^j \bigg]$$
$\alpha$ isn't necessarily bigger than $\beta$ or vice versa
In general when multiplying two sums, you get: $$\left(\sum_{k=0}^{n}a_k\right) \cdot \left(\sum_{j=0}^{m} b_j\right) = \sum_{k=0}^{n}a_k \cdot {\left(\sum_{j=0}^{m} b_j\right)} = \sum_{k=0}^{n}\sum_{j=0}^{m}a_kb_j$$
In your case it would be: $$a_k = \binom{\alpha+1}{k} \cdot x{_{i}}^{\alpha-k+1} \cdot (x_{i+1}-x_{i})^k \cdot t^k$$
with $n = \alpha +1$ and $$b_j =\binom{\beta}{j} \cdot y{_{i}}^{\beta-j} \cdot (y_{i+1}-y_{i})^j \cdot t^j$$ with $m = \beta$
I belive, you can take it from there.