ホッケースティック恒等式

ホッケースティック恒等式:

• ${\sum }_{i=0}^k\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)=\left(\genfrac{}{}{0ex}{}{n+k+1}{k}\right)$

• proof

  • 負の二項定理 eq4-1

    • $1/(1-x{)}^{d+1}={\sum }_{n=0}^{\infty }(\genfrac{}{}{0ex}{}{n+d}{d})x^n$

  • 形式的べき級数の係数の部分和 eq4-2

    • $\sum _{i=0}^k[x^i]F=[x^k]\frac{1}{1-x}F$

  • eq4-3

    • $\left(\genfrac{}{}{0ex}{}{n+m}{m}\right)=\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$

  • ${\sum }_{i=0}^k\left(\genfrac{}{}{0ex}{}{n+i}{i}\right)={\sum }_{i=0}^k[x^i]{\sum }_{m=0}^{\infty }\left(\genfrac{}{}{0ex}{}{n+m}{m}\right)x^m$ … 係数の部分和

  • ${\sum }_{i=0}^k[x^i]{\sum }_{m=0}^{\infty }\left(\genfrac{}{}{0ex}{}{n+m}{m}\right)x^m=[x^k]\frac{1}{1-x}{\sum }_{m=0}^{\infty }\left(\genfrac{}{}{0ex}{}{n+m}{m}\right)x^m$ … by eq4-2

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

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

  • $[x^k]\frac{1}{1-x}\frac{1}{(1-x{)}^{n+1}}=[x^k]\frac{1}{(1-x{)}^{n+2}}$ … 整理

  • $[x^k]\frac{1}{(1-x{)}^{n+2}}=[x^k]{\sum }_{m=0}^{\infty }\left(\genfrac{}{}{0ex}{}{m+n+1}{n+1}\right)x^n$ … by eq4-1

  • $[x^k]{\sum }_{m=0}^{\infty }\left(\genfrac{}{}{0ex}{}{m+n+1}{n+1}\right)x^n=\left(\genfrac{}{}{0ex}{}{k+n+1}{n+1}\right)$ … 形式的べき級数の係数

  • $\left(\genfrac{}{}{0ex}{}{k+n+1}{n+1}\right)=\left(\genfrac{}{}{0ex}{}{k+n+1}{k}\right)$ … by eq4-3

  • パスカルの三角形的解釈

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    • ホッケースティック恒等式|思考力を鍛える数学

二項係数の公式