eq6-3をeq6-2に帰着

from 二項係数の公式 eq6-3をeq6-2に帰着 - ${\sum }_i\left(\genfrac{}{}{0ex}{}{A+i}{i}\right)\left(\genfrac{}{}{0ex}{}{B+K-i}{K-i}\right)$ - $={\sum }_i\left(\genfrac{}{}{0ex}{}{A+i}{A}\right)\left(\genfrac{}{}{0ex}{}{B+K-i}{B}\right)$ … eq4-3 - $={\sum }_j\left(\genfrac{}{}{0ex}{}{j}{A}\right)\left(\genfrac{}{}{0ex}{}{B+K-j+A}{B}\right)$ … A+i -> j - $={\sum }_j\left(\genfrac{}{}{0ex}{}{j}{A}\right)\left(\genfrac{}{}{0ex}{}{N-j-1}{B}\right)$ … A+B+K+1 -> N - $=\left(\genfrac{}{}{0ex}{}{N}{A+B+1}\right)$ … eq6-2 - $=\left(\genfrac{}{}{0ex}{}{A+B+K+1}{A+B+1}\right)=\left(\genfrac{}{}{0ex}{}{A+B+K+1}{K}\right)$