|
ZCos666
Posted at 2025-3-18 19:05:00
第三个就是第一个,第二个也包含在第一个中
\[ \begin{aligned}
\sum_{k=0}^{r-1}(-1)^{r+k+1}\binom{2r}{k}&=(-1)^{r+1}\cdot\dfrac{(-1)^{r+1}}{2}\binom{2r}{r}\\
&=\dfrac{1}{2}\binom{2r}{r}
\end{aligned} \]
\[ \begin{aligned}
\sum_{k=0}^{r-1}\left((-1)^{r-k+1}-1\right)\binom{2r}{k}&=\dfrac{1}{2}\binom{2r}{r}-\sum_{k=0}^{r-1}\binom{2r}{k}\\
&=\dfrac{1}{2}\binom{2r}{r}-\dfrac{1}{2}\left(4^r-\binom{2r}{r}\right)\\
&=\binom{2r}{r}-2^{2r-1}
\end{aligned} \]
|
Rate
-
View Rating Log
|