@[Link_Cut_Y](/user/519384) 这个就是你先把 $\frac{1}{(1-x)^4}$ 展开成 $\binom{3}{0}+\binom{3+1}{1}x+\binom{3+2}{2}x^2+\cdots$，然后乘个 $x$ 就行了（即系数整体右移一位）。
by Yusani_huh @ 2024-02-22 18:40:34
@[Yusani_huh](/user/239895) emm，这里是怎么展开的能不能细说一下ww
by Link_Cut_Y @ 2024-02-22 18:43:42
展开为啥是那样的的证明大概就是
$$
\begin{aligned}
(1-x)^{-4}&=\sum_{i=0}\frac{(-4)^{\underline i}}{i!}(-x)^i\\
&=\sum_{i=0}\frac{(-4)^{\underline i}(-1)^i}{i!}x^i\\
&=\sum_{i=0}\frac{(4+i-1)^{\underline i}}{i!}x^i\\
&=\sum_{i=0}\binom{3+i}{i}x^i\\
\end{aligned}
$$
by Yusani_huh @ 2024-02-22 18:45:48
@[Link_Cut_Y](/user/519384)
by Yusani_huh @ 2024-02-22 18:45:58
$$
\begin{aligned}
\frac{x}{(1-x)^{4}} & =x(1-x)^{-4}\\
& =x\sum_{k=0}^{\infty}\binom{-4}{k}(-x)^{k}1^{-4-k}\\
& =x\sum_{k=0}^{\infty}(-1)^{k}\binom{k-(-4)-1}{k}(-x)^{k}\\
& =x\sum_{k=0}^{\infty}\binom{k+3}{3}x^{k}\\
& =\sum_{k=0}^{\infty}\binom{k+3}{3}x^{k+1}\\
& =\sum_{k=1}^{\infty}\binom{k+2}{3}x^{k}
\end{aligned}
$$
by _saltFish_ @ 2024-02-22 18:46:50
@[Yusani_huh](/user/239895) 您的式子貌似有点问题。
by _saltFish_ @ 2024-02-22 18:47:48
@[Link_Cut_Y](/user/519384)
by _saltFish_ @ 2024-02-22 18:48:04
@[Yusani_huh](/user/239895) 第二行到第三行那里是把下降幂转成上升幂，然后推过来了吗/yiw
by Link_Cut_Y @ 2024-02-22 18:49:02
@[_saltFish_](/user/661044) 没问题吧，咱俩结果不是一样的吗。
@[Link_Cut_Y](/user/519384) 那一步是把下降幂展开然后每一项都乘上 $-1$
by Yusani_huh @ 2024-02-22 18:50:56
@[Yusani_huh](/user/239895) 哦哦我会了
谢谢神！
by Link_Cut_Y @ 2024-02-22 18:51:55