<s>[INST] Can we express the following ordinary generating function? I wish to express the following power series 
$$ \sum_{k \ge 0} \binom{n-k}{m} x^k$$
where $n,m$ are positive integer such that  $0< m \le n$
 [/INST] In fact, the polynomial part $$\sum_{k=0}^n \binom{n-k}{m} x^k= \sum_{k=0}^{n-m} \binom{n-k}{m} x^k + \sum_{k=n-m+1}^n \binom{n-k}{m} x^k$$ 
$$=\sum_{k=0}^{n-m} \binom{n-k}{m} x^k$$
$$ =x^{n} \sum_{k=0}^{n-m} \binom{n-k}{m} x^{k-n} $$ $$=x^{n-m} [\sum_{i=0}^{\infty} \binom{m+i}{m} \frac{1}{x^i} - \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}]$$ 
$$ =x^{n-m} \frac{1}{(1- (1/x))^{m+1}} - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i} $$
$$ = \frac{x^{n+1}}{(x- 1)^{m+1}} - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}.$$
Hence,  $$\sum_{k=0}^{\infty} \binom{n-k}{m} x^k = - x^{n-m} \sum_{i=n-m+1}^{\infty} \binom{m+i}{m} \frac{1}{x^i}.$$
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