ARC110D

Solve ARC110D by attributing it to [formal power series

$F_{i} (x) = \sum_{k = 0}^{\infty} (A _{i} k) x^{k}$ $G (x) = \prod_{i = 1}^{N} F_{i} (x)$ - Lower fixed binomial coefficient → negative binomial theorem : - $(A A + i) = [x^{i}] \sum_{n} (A A + n) x^{n} = [x^{i}] \frac{1}{( 1 - x ) ^{A + 1}}$

$i < 0$ , since $(A A + i) = 0$ $F_{i} (x) = \sum_{k = 0}^{\infty} (A _{i} k) x^{k} = \frac{x ^{A_{i}}}{( 1 - x ) ^{A_{i} + 1}}$

$S := \sum A_{i}$

$G (x) = \prod_{i = 1}^{N} F_{i} (x) = \frac{x ^{S}}{( 1 - x ) ^{S + N}}$

- [[Partial sum of coefficients of formal power series]] :

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

$X = \sum_{i =}^{M} [x^{i}] G = [x^{M}] \frac{1}{1 - x} G = [x^{M}] \frac{x ^{S}}{( 1 - x ) ^{S + N + 1}}$ $= [x^{M - S}] \frac{1}{( 1 - x ) ^{S + N + 1}}$

- [[Power series → binomial coefficient]] :
- $[x^B]\frac{1}{(1-x)^{A}} = \binom{A+B-1}{A-1}$

$[x^{M - S}] \frac{1}{( 1 - x ) ^{S + N + 1}} = (S + N S + N + 1 + M - S - 1) = (S + N N + M)$

reference

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ARC110D_FPS

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