下固定の二項係数→負の二項定理
\binom{A+i}{A} = [x^i] \sum_n \binom{A+n}{A}x^n = [x^i]\frac{1}{(1-x)^{A+1}}
(1-x)^{-d} = \sum_{n=0}^\infty \binom{n+d-1}{d-1}x^n
逆向き
[x^B]\frac{1}{(1-x)^{A}} = \binom{A+B-1}{A-1}
関連
これは母関数で
Related Pages
"Engineer's way of creating knowledge" the English version of my book is now available on [Engineer's way of creating knowledge]
(C)NISHIO Hirokazu / Converted from [Scrapbox] at [Edit]