eq7-proof

from binomial coefficient formula

• eq7-proof

  • Consider a formal power series F such that $[x^k]F={\sum }_{i=0}^k\left(\genfrac{}{}{0ex}{}{n+1}{i}\right)\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)$

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

  • When $k<i$ $\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)=0$, so ${\sum }_{i=0}^k$ can be [$ \sum_{i=0}^{\infty}

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

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

  • $F=\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)\sum _{k=0}^{\infty }\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)x^k\right)$ … We can dwell on the coefficients independent of k, and then use

  • $k<i$ as $\left(\genfrac{}{}{0ex}{}{m-i}{k-i}\right)=0$ so [$ j=k - i \quad (j > 0)

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

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

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

  • $F=(1+x{)}^m\sum _{i=0}^{\infty }\left(\left(\genfrac{}{}{0ex}{}{n+i}{i}\right){\left(\frac{x}{1+x}\right)}^i\right)$ … We can lump together the coefficients independent of i

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

  • $F=(1+x{)}^m/{\left(1-\frac{x}{1+x}\right)}^{n+1}$ … negative binomial theorem

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

  • $F=(1+x{)}^m(1+x{)}^{n+1}$

  • $F=(1+x{)}^{m+n+1}$

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

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

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