eq7-proof
eq7-proof
Consider a formal power series F such that $ [x^k]F = \sum_{i=0}^k \binom{n+1}{i}\binom{m-i}{k - i}
$ \displaystyle F = \sum _ {k=0}^{\infty} \sum_{i=0}^{k} {n+i \choose i}{m-i \choose k-i} x^{k}
When $ k < i $ \binom{m-i}{k-i} = 0, so $ \sum_{i=0}^{k} can be [$ \sum_{i=0}^{\infty}
$ \displaystyle F = \sum _ {k=0}^{\infty} \sum _ {i=0}^{\infty} {n+i \choose i}{m-i \choose k-i} x^{k}
$ \displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} \sum_{k=0}^{\infty} {m-i \choose k-i} x^{k}\right) ... We can dwell on the coefficients independent of k, and then use
$ k < i as $ \binom{m-i}{k-i} = 0 so [$ j=k - i \quad (j > 0)
$ \displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} \sum_{j=0}^{\infty} {m-i \choose j} x^{j}x^i\right)
$ \displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} (1+x)^{m-i} x^i\right) ... binomial theorem $ \displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} (1+x)^m(1+x)^{-i} x^i\right)
$ \displaystyle F = (1+x)^m \sum_{i=0}^{\infty}\left({n+i \choose i} \left(\frac{x}{1+x}\right)^i \right) ... We can lump together the coefficients independent of i
$ \displaystyle F = (1+x)^m \sum_{i=0}^{\infty}\left({i+n \choose n} \left(\frac{x}{1+x}\right)^i \right) ... by eq4-3
$ \displaystyle F = (1+x)^m / \left(\frac{1}{1+x}\right)^{n+1}
$ \displaystyle F = (1+x)^m (1+x)^{n+1}
$ \displaystyle F = (1+x)^{m+n+1}
$ [x^k]F = \binom{m+n+1}{k}
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