$(2n-1)!!$的恒等式

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A Combinatorial Survey of Identities for the Double Factorial.pdf (235.62 KB, Downloads: 52)

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4.1$$\sum_{k=0}^{n-1}\left(\begin{array}{c}n \\ k+1\end{array}\right)(2 k-1) ! !(2 n-2 k-3) ! !=(2 n-1) ! !$$
4.2$$\sum_{k=0}^{n}\left(\begin{array}{c}2 n-k-1 \\ k-1\end{array}\right) \frac{(2 n-2 k-1)(2 n-k+1)}{k+1}(2 n-2 k-3) ! !=(2 n-1) ! !$$
4.3$$\sum_{k=1}^{n} \frac{(n-1) !}{(k-1) !} k(2 k-3) ! !=(2 n-1) ! !$$
4.4$$\sum_{k=0}^{n / 2}\left(\begin{array}{c}n \\ 2 k\end{array}\right)\left(\begin{array}{c}2 k \\ k\end{array}\right) \frac{n !}{2^{2 k}}=(2 n-1) ! !$$
4.5$$\sum_{k=1}^{n} \frac{(2 n-2) ! !(2 k-3) ! !}{(2 k-2) ! !}=(2 n-1) ! !$$
4.6$$(2 n-2) ! !+\sum_{k=2}^{n} \frac{(2 n-1) ! !(2 k-4) ! !}{(2 k-1) ! !}=(2 n-1) ! !$$