i want to change the lower bound of the summation with $k=0$. $$\sum_{k=2}^{r+s-1} { \{ \begin{pmatrix} k-1\\r-1\\ \end{pmatrix} +\begin{pmatrix} k-1\\s-1\\ \end{pmatrix} \} } H(k,r+s-k) $$
In here, I want to take the sum from $k=0$. $$\sum_{k=2-\color{red}{2}}^{r+s-1-\color{red}{2}} $$.
Then, how can I change the middle part?
I know to change the bounderies in a summation like below $$\sum_{i=x}^{y} f(i)=\sum_{i=x-a}^{y-a} f(i+a) $$.
And also there is another property $$\sum_{i=x}^{y} f(i,y)=\sum_{i=x+t}^{y+t} f(i-t,y-2t) $$.
But sometimes this doesn't work. Could you please explain this also?
One way to do it is to make a formal substitution $\ell=k-2$, so that $k=\ell+2$. Then
$$\sum_{k=2}^{r+s-1}\left(\binom{k-1}{r-1}+\binom{k-1}{s-1}\right)H(k,r+s-k)$$
automatically becomes
$$\sum_{\ell=0}^{r+s-3}\left(\binom{\ell+1}{r-1}+\binom{\ell+1}{s-1}\right)H(\ell+2,r+s-\ell-2)\;.$$
If you wish, you can now simply rename $\ell$ back to $k$ to get
$$\sum_{k=0}^{r+s-3}\left(\binom{k+1}{r-1}+\binom{k+1}{s-1}\right)H(k+2,r+s-k-2)\;.$$
As a quick partial check, note that the upper number of the binomial coefficients is still $1$ less than the first argument of $H$, as it should be, and the two arguments of $H$ still sum to $r+s$, as they should.
Added to cover the addition to the question: Suppose that we start with the summation
$$\sum_{i=x}^yf(i,y)$$
and let $j=i+t$, so that $i=j-t$. Making the substitution, we see that
$$\sum_{i=x}^yf(i,y)=\sum_{j=x+t}^{y+t}f(j-t,y)\;,$$
and we can then rename $j$ back to $i$ to find that
$$\sum_{i=x}^yf(i,y)=\sum_{i=x+t}^{y+t}f(i-t,y)\;.$$
There is no change in the second argument of $f$, because $y$ is a constant: it does not depend in any way on the index variable. The range of values of the index variable depends on $y$, but that’s a different thing altogether. The formula
$$\sum_{i=x}^{y} f(i,y)=\sum_{i=x+t}^{y+t} f(i-t,y-2t)$$
is simply wrong.