how can be prove expression:
$\sum \limits_{s = 0}^{2k} (-1)^s\binom{n+s}{n}\binom{n+2k-s}{n} = \binom{n+k}{k}$
by using this identity:
$(1 − t)^{−n−1}(1 + t)^{−n−1}= (1 − t^2)^{−n−1},$
or how can be prove expression:
$\sum \limits_{s = 0}^{k} (-1)^{k-s}\binom{n}{s}\binom{m+k-s-1}{k-s} =\binom{n-m}{k}$
by using this identity:
$(1 + t)^n (1 + t)^{−m} = (1 + t)^{n−m}.$
I open brackets by Binomial Theorem, but what kind of coefficients i should equate?
(Second problem)
Using Upper Negation, followed by the Vandermonde Identity:
$$\begin{align} \sum_{s=0}^{k}(-1)^{k-s}{n \choose s}{m+k-s-1\choose k-s}&=\sum_{s=0}^{k}(-1)^{k-s}{n \choose s}{(-m+1)-1\choose k-s}(-1)^{k-s}\\ &=\sum_{s=0}^{k}\underbrace{(-1)^{2(k-s)}}_{1}{n \choose s}{-m\choose k-s}\\ &={n-m\choose k}\qquad \blacksquare \end{align}$$