eq7-proof

from 二項係数の公式

• eq7-proof

  • $[x^k]F={\sum }_{i=0}^k\left(\genfrac{}{}{0ex}{}{n+1}{i}\right)\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)$であるような形式的べき級数Fを考える

  • $F=\sum _{k=0}^{\infty }\sum _{i=0}^k\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)x^k$

  • $k<i$の時$\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)=0$ なので ${\sum }_{i=0}^k$を ${\sum }_{i=0}^{\infty }$ にして良い

  • $F=\sum _{k=0}^{\infty }\sum _{i=0}^{\infty }\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)x^k$

  • $F=\sum _{i=0}^{\infty }\sum _{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)x^k$ … 足し算の順序の変更

  • $F=\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)\sum _{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)x^k\right)$ … kに依存しない係数をくくりだし

  • $k<i$の時$\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)=0$ なので$j=k-i(j>0)$として

  • $F=\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)\sum _{j=0}^{\infty }\left(\genfrac{}{}{0ex}{}{m-i}{j}\right)x^j{x}^i\right)$

  • $F=\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)(1+x{)}^{m-i}{x}^i\right)$ … 二項定理

  • $F=\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)(1+x{)}^m(1+x{)}^{-i}{x}^i\right)$

  • $F=(1+x{)}^m\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{n+i}{i}\right){\left(\frac{x}{1+x}\right)}^i\right)$ … iに依存しない係数をくくりだし

  • $F=(1+x{)}^m\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{i+n}{n}\right){\left(\frac{x}{1+x}\right)}^i\right)$ … by eq4-3

  • $F=(1+x{)}^m/{\left(1-\frac{x}{1+x}\right)}^{n+1}$ … 負の二項定理

  • $F=(1+x{)}^m/{\left(\frac{1}{1+x}\right)}^{n+1}$

  • $F=(1+x{)}^m(1+x{)}^{n+1}$

  • $F=(1+x{)}^{m+n+1}$

  • $F=\sum _{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{m+n+1}{k}\right)x^k$… 二項定理

  • $[x^k]F=\left(\genfrac{}{}{0ex}{}{m+n+1}{k}\right)$