eq7-proof

NISHIO Hirokazu [Translate]

eq7-proof
from 二項係数の公式
eq7-proof
[x^k]F = \sum_{i=0}^k \binom{n+1}{i}\binom{m-i}{k - i}であるような形式的べき級数Fを考える
\displaystyle F = \sum _ {k=0}^{\infty} \sum_{i=0}^{k} {n+i \choose i}{m-i \choose k-i} x^{k}
k < iの時\binom{m-i}{k-i} = 0 なので \sum_{i=0}^{k}を \sum_{i=0}^{\infty} にして良い
\displaystyle F = \sum _ {k=0}^{\infty} \sum _ {i=0}^{\infty} {n+i \choose i}{m-i \choose k-i} x^{k}
\displaystyle F = \sum_{i=0}^{\infty} \sum_{k=0}^{\infty} {n+i \choose i}{m-i \choose k-i} x^{k} ... 足し算の順序の変更
\displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} \sum_{k=0}^{\infty} {m-i \choose k-i} x^{k}\right) ... kに依存しない係数をくくりだし
k < iの時\binom{m-i}{k-i} = 0 なのでj=k - i \quad (j > 0)として
\displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} \sum_{j=0}^{\infty} {m-i \choose j} x^{j}x^i\right)

\displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} (1+x)^{m-i} x^i\right) ... 二項定理

\displaystyle F = \sum_{i=0}^{\infty}\left({n+i \choose i} (1+x)^m(1+x)^{-i} x^i\right)
\displaystyle F = (1+x)^m \sum_{i=0}^{\infty}\left({n+i \choose i} \left(\frac{x}{1+x}\right)^i \right) ... iに依存しない係数をくくりだし
\displaystyle F = (1+x)^m \sum_{i=0}^{\infty}\left({i+n \choose n} \left(\frac{x}{1+x}\right)^i \right) ... by eq4-3
\displaystyle F = (1+x)^m / \left(1 - \frac{x}{1+x}\right)^{n+1}  ... 負の二項定理
\displaystyle F = (1+x)^m / \left(\frac{1}{1+x}\right)^{n+1} 

\displaystyle F = (1+x)^m (1+x)^{n+1} 

\displaystyle F = (1+x)^{m+n+1} 
\displaystyle F = \sum_{k=0}^\infty \binom{m+n+1}{k} x^k... 二項定理
[x^k]F = \binom{m+n+1}{k}

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(C)NISHIO Hirokazu / Converted from [Scrapbox] at 11/23/2025, 6:17:33 PM[Edit]

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