How does this summation hold?

Question

How does $$ \sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} = \sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k} $$ hold?

RobJohn helped me, but I could do only this: $$\sum_{r=0}^{m+n} \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{r} \ x^{r+k}=\sum_{r=0}^{n} \sum_{k=0}^{m} \binom{n}{k} \binom{m}{r} \ x^{r+k}.$$ Please help!

Answer 1

In general $$\sum_{k=0}^n\sum_{j=0}^k=\sum_{j=0}^k\sum_{k=j}^n$$

Now in your case we can apply this to get $$\sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} $$ equals $$\sum_{k=0}^{r} \left( \sum_{r=k}^{m+n} \binom{n}{k} \binom{m}{r-k} x^{r-k} \right) \ x^{k} $$

This is the same as $$\sum_{k=0}^{r} \left( \sum_{r=0}^{m+n-k} \binom{n}{k} \binom{m}{r} x^r\right) \ x^k$$

Can you move on?

Answer 2

Consider \begin{align} S = \sum_{r=0}^{m+n} \left( \sum_{k=0}^{r} \binom{n}{k} \binom{m}{r-k} \right) \ x^{r} \end{align} and shift the index $r$ to $r = r+k$ to obtain \begin{align} S = \sum_{r=0}^{m+n} \left( \sum_{k=0}^{m+n} \binom{n}{k} \binom{m}{k} \right) \ x^{r+k} \end{align} Now both sides are equal.