Binomial Coefficient Summation Equivalence 
Following on a previous question, I'm supposed to prove the following:
  $$ \sum_{i=0}^{n-1} \binom{2n-2-i}{n-1} =  \sum_{i=0}^{n-1} \binom{n-1+i}{i}.$$
  Is there any simple conversion to come from the first term to the second one?

 A: A slightly neater way to write what you need to show (letting $N=n-1$) is
$$\color{green}{\sum_{i=0}^{N} \binom{2N-i}{N}} =  \color{blue}{\sum_{i=0}^{N} \binom{N+i}{i}}.$$
Using the identity ${m\choose k}\color{red}={m\choose m-k},$
$$\color{blue}{\sum_{i=0}^{N} \binom{N+i}i} \color{red}= \sum_{i=0}^{N} \binom{N+i}{N}=\sum_{M=N}^{2N} \binom{M}{N}=\sum_{M=2N-N}^{2N-0} \binom{M}{N}=\color{green}{\sum_{i=0}^{N} \binom{2N-i}{N}}.$$
A: Note that $$\binom{m}{i}=\binom{m}{m-i}.$$
Therefore,
$$\binom{2n-2-i}{n-1} = \binom{2n-2-i}{2n-2-i-n+1}=\binom{2n-2-i}{n-1-i}.$$
Consequently,
$$\sum_{i=0}^{n-1}\binom{2n-2-i}{n-1} = \sum_{i=0}^{n-1}\binom{2n-2-i}{n-1-i} = \binom{2n-2}{n-1}+\binom{2n-3}{n-2}+\cdots+\binom{n-1}{0}.$$
That is
$$\sum_{i=0}^{n-1}\binom{2n-2-i}{n-1} =  \sum_{i=0}^{n-1}\binom{n-1+i}{i}.$$
A: Since you ask for a "transformation", then consider that
$$ \bbox[lightyellow] {  
\eqalign{
  & \sum\limits_{0\, \le \,i\, \le \,n - 1} {
\left( \matrix{  2n - 2 - i \cr   n - 1 \cr}  \right)}  =    \quad \quad (a.0)  \cr 
  &  = \sum\limits_{0\, \le \,i\, \le \,n - 1} {
\left( \matrix{  2n - 2 - i \cr   n - 1 - i \cr}  \right)  }  =   \quad \quad (a.1)   \cr 
  &  = \sum\limits_{0\, \le \,i\,\left( { \le \,n - 1} \right)} {
  \left( \matrix{2n - 2 - i \cr   n - 1 - i \cr}  \right)}  =    \quad \quad (a.2)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n - 1} \right)} {
\left( \matrix{  2n - 2 - i \cr   n - 1 - i \cr}  \right) \left( \matrix{  i \hfill \cr   i \hfill \cr}  \right)}  =   \quad \quad (a.3)   \cr 
  &  = \left( \matrix{  2n - 1 \cr n - 1 \cr}  \right)   \quad \quad (a.4) \cr} 
}$$
and
$$ \bbox[lightyellow] {  
\eqalign{
  & \sum\limits_{0\, \le \,i\, \le \,n - 1} {
  \left( \matrix{n - 1 + i \cr  i \cr}  \right)  }  =   \quad \quad (b.0) \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,i\, \le \,n - 1} {
 \left( \matrix{ n - 1 + i \cr   i \cr}  \right)   }  =   \quad \quad (b.1)  \cr 
  &  = \sum\limits_{\left( {0\, \le } \right)\,i\,\left( { \le \,n - 1} \right)} {
\left( \matrix{  n - 1 + i \cr   i \cr}  \right)\left( \matrix{  n - 1 - i \cr   n - 1 - i \cr}  \right)  }  =    \quad \quad (b.2) \cr 
  &  = \left( \matrix{  2n - 1 \cr   n - 1 \cr}  \right)   \quad \quad (b.3)\cr} 
}$$
where:
 - a.1)  symmetry : we can do that because the upper term is non-negative for $0 \le i \le n-1$;
 - a.2)  upper bound in brackets to mean that we can remove it since it is implicit in the binomial;
 - a.3)  lower bound in brackets: it is replaced by the additional binomial;
 - a.4)  double convolution, we can apply it since the index is free to vary over all the allowed range;
 - b.1)  lower bound in brackets : it is implicit in the binomial$;
 - b.2)  upper bound in brackets: replaced by the second binomial;
 - b.3)  double convolution: the index is free to vary and we can apply it;   
A: 
We obtain
  \begin{align*}
\color{blue}{\sum_{i=0}^{n-1}\binom{2n-2-i}{n-1}}
&=\sum_{i=0}^{n-1}\binom{2n-2-i}{n-1-i}\tag{1}\\
&\,\,\color{blue}{=\sum_{i=0}^{n-1}\binom{n-1+i}{i}}\tag{2}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

*In (2) we change the order of summation $i\to n-1-i$.
