Water__Problem @ 2025-01-24 10:52:19
一些组合恒等式不会证：
1.\sum^n_{k=0}kC^{n-k}_{2*n+1}=(2n+1)C^n_{2n-1}-2^{2n-1}
2.\sum^n_{k=0}kC^{n-k}_{2n}=nC^n_{2n-1}
3.\sum^{2n-1}_{k=1}(-1)^{k-1} \frac{n-k}{C^k_{2n}}=0
必关！！！
by liujiaxi123456 @ 2025-01-24 11:48:58
@Water__Problem
左式 = \sum_{k=0}^n (n+k)\cdot C^{n+k}_{2n} - n\cdot C_{2n}^{n+k}
运用吸收公式，得到：
\sum 2n\cdot C_{2n-1}^{n+k-1} - n\cdot C^{n+k}_{2n}
令 S(n, m) = \sum_{i=1}^m C^i_n
= 2n\cdot S(2n-1, 2n) - n\cdot S(2n, 2n) - 2n\cdot S(2n-1, n-2) + n\cdot S(2n, n-1)
因为：
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S(n, m+1) = S(n, m) + C_n^{m+1}
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S(n+1, m) = 2\cdot S(n, m) - C_n^m
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S(n, m) = \frac{S(n+1, m+1) - C_{n}^{m+1}}{2}
所以：
= n\cdot (S(2n, 2n)-C^{2n}_{2n-1}) - n\cdot S(2n, 2n) - n\cdot (S(2n, n-1)-C^{n-1}_{2n-1}) + n\cdot S(2n, n-1)
= -n\cdot C_{2n-1}^{2n} + n\cdot C_{2n-1}^{n-1} = n\cdot C_{2n-1}^{n-1} = n\cdot C_{2n-1}^n
by liujiaxi123456 @ 2025-01-24 12:07:41
@Water__Problem
令 S_1 = \sum_{k=1}^{2n-1} (-1)^{k-1} \frac{n}{\binom{2n}{k}} = n \sum_{k=1}^{2n-1} (-1)^{k-1} \frac{1}{\binom{2n}{k}}
令 S_2 = \sum_{k=1}^{2n-1} (-1)^{k-1} \frac{k}{\binom{2n}{k}}
考虑对 k 进行换元，使 k' = 2n-k
则 S_2 = \sum_{k=1}^{2n-1} (-1)^{k-1} \frac{2n-k}{\binom{2n}{2n-k}} = \sum_{k=1}^{2n-1} (-1)^{k-1} \frac{2n-k}{\binom{2n}{k}} = 2S_1 - S_2
由此得证：S_1 = S_2
证毕
by liujiaxi123456 @ 2025-01-24 15:07:56
@Water__Problem
左式 = \sum_{k=0}^n (n-k)\cdot C^k_{2n+1} = n\sum_{k=0}^n C_{2n+1}^k - \sum_{k=0}^n k\cdot C_{2n+1}^k
两个拆开处理，分别为 s1,s2：
S_1 = n\cdot (\sum_{k=0}^{2n+1}C_{2n+1}^k - \sum_{k=0}^{n+1}C_{2n+1}^{n+1+k})
= n\cdot (\sum_{k=0}^{2n+1}C_{2n+1}^k - \sum_{k=0}^{n}C_{2n+1}^{k})
\therefore 2S_1 = n\cdot \sum_{k=0}^{2n+1} C_{2n+1}^k = n\cdot 2^{2n+1}
\therefore S_1 = n\cdot 2^{2n}
S_2 = \sum_{k=0}^n(2n+1)\cdot C_{2n}^{k-1} = (2n+1)\sum_{k=0}^{n-1}C_{2n}^k
= (2n+1)(\sum_{k=0}^{2n}C_{2n}^k - \sum_{k=0}^{n-1}C_{2n}^k - C_{2n}^n)
\therefore \sum_{k=0}^{n-1}C_{2n}^k = 2^{2n-1} - \frac{C_2n^n}{2}
\therefore S_2 = n\cdot 2^{2n} + 2^{2n-1} - n\cdot C_{2n}^n - \frac{C_{2n}^n}{2}
S_1 - S_2 = (2n+1)\cdot \frac{C_{2n}^n}{2} - 2^{2n-1}
和右式比较一下，就是要证 C_{2n}^n = 2C_{2n-1}^n
\because C_{2n}^n = C_{2n-1}^n + C_{2n-1}^{n-1}
所以需证： C_{2n-1}^n = C_{2n-1}^{n-1}
而由于组合数有对称性，所以显然成立。