$$\mathbb{X} \sim BN(m,p)$$
$$\mathbb{Y} \sim BN(n,p)$$
Sea \(\mathhbb{Z}=\mathbb{X}+\mathbb{Y}\), queremos ver que \(\mathbb{Z}\sim BN(m+n,p)\) .
$$P(\mathbb{Z}=j)=\sum_{k=0}^j P(\mathbb{Y}=j-k)P(\mathbb{X}=k)=$$
$$= \sum_{k=0}^j \binom{j-k+n-1}{n-1}.(1-p)^n . p^{j-k} . \binom{k+m-1}{m-1} . (1-p)^m . p^k=$$
$$=\sum_{k=0}^j \binom{j-k+n-1}{n-1} \binom{k+m-1}{m-1}.(1-p)^{m+n} . p^j=$$
Ahora miramos:
$$\binom{k+r-1}{r-1}=(-1)^k \binom{-r}{k}$$
Por lo tanto
$$\binom{j-k+n-1}{n-1}=(-1)^{j-k} * \binom{-n}{j-k}$$
$$\binom{k+m-1}{m-1}=(-1)^k \binom{-m}{k}$$
Y asi:
$$\binom{j-k+n-1}{n-1} \binom{k+m-1}{m-1} = (-1)^j \binom{-m}{k} \binom{-n}{j-k}$$
$$(-1)^j \sum_{k=0}^j \binom{-m}{k} \binom{-n}{j-k} = (-1)^j \binom{-(m+n)}{j}$$
$$=\binom{j+m+n-1}{m+n-1}$$
Luego:
$$=\sum_{k=0}^j \binom{j-k+n-1}{n-1} \binom{k+m-1}{m-1}.(1-p)^{m+n} . p^j=$$
$$=\binom{j+m+n-1}{m+n-1}(1-p)^{m+n} . p^j=P(\mathbb{Z}=j)$$
Por lo tanto:
$$\mathbb{Z}\sim BN(m+n,p)$$