from binomial coefficient formula
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}
$\displaystyle F = \sum _ {k=0}^{\infty} \sum _ {i=0}^{\infty} {n+i \choose i}{m-i \choose k-i} x^{k}$
$\displaystyle F = \sum_{i=0}^{\infty} \sum_{k=0}^{\infty} {n+i \choose i}{m-i \choose k-i} x^{k}$ ... Change the order of addition
$\displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} \sum_{k=0}^{\infty} {m-i \choose k-i} x^{k}\right)$ ... 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(1 - \frac{x}{1+x}\right)^{n+1} $ ... negative binomial theorem
$\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} $
$\displaystyle F = \sum_{k=0}^\infty \binom{m+n+1}{k} x^k$... binomial theorem
$[x^k]F = \binom{m+n+1}{k}$
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