组合数学第五章练习题（部分）
组合数学第五章练习题（部分）
11.
\[\binom{n}{k} - \binom{n - 3}{k} = \binom{n - 1}{k - 1} + \binom{n - 2}{k - 1} + \binom{n - 3}{k - 1}
\]
理树要在神北私立高级中学的 \(n\) 位女同学中挑选 \(k\) 位后宫，但是他必须走沙耶，佳奈多和撒撒撒撒撒米中的至少一条线，因为这是 LBEX。
12.
\[\begin{aligned}
\sum\limits_{k = 0}^{n}(-1)^k\binom{n}{k}^2 =
\begin{cases}
0, & \text{若 } n \text{ 是奇数} \\
(-1)^m\binom{2m}{m}, & \text{若 } n = 2m
\end{cases}
\end{aligned}
\]
\[\begin{aligned}
(1 - x^2)^n &= (1 + x)^n(1 - x)^n \\
\sum\limits_{i = 0}^{n}\binom{n}{i}(-x^2)^{i} &= \left( \sum\limits_{i = 0}^{n}\binom{n}{i}x^i \right)\left( \sum\limits_{i = 0}^{n}\binom{n}{i}(-x)^i \right) \\
\sum\limits_{i = 0}^{n}\binom{n}{i}(-1)^ix^{2i} &= \sum\limits_{i = 0}^{2n}\sum\limits_{k = 0}^{i}(-1)^k\binom{i}{k}\binom{i}{i - k}x^i \\
\end{aligned}
\]
原式显然。这道卷积应该是前后几道题中最有意思的一道。
13.
\[\begin{aligned}
ans = &\binom{n}{k} + 3\binom{n}{k - 1} + 3\binom{n}{k - 2} + \binom{n}{k - 3} \\
&= \sum\limits_{i = 0}^{3}\binom{3}{i}\binom{n}{k - i} = \binom{n + 3}{k}
\end{aligned}
\]
14.
\[\begin{aligned}
\frac{r}{r - k}\binom{r - 1}{k} = \frac{r}{r - k}\times \frac{(r - 1)!}{(r - k - 1)!k!} = \frac{r!}{(r - k)!k!} = \binom{r}{k}
\end{aligned}
\]
15.
\[\sum\limits_{i = 1}^{n}(-1)^{i + 1}i\binom{n}{i} = \sum\limits_{i = 1}^{n}(-1)^{i + 1}n\frac{(n - 1)!}{(i - 1)!(n - i)!} = n\sum\limits_{i = 0}^{n - 1}(-1)^i\binom{n - 1}{i}\times 1^{n - i} = 0
\]
16.
\[\begin{aligned}
(1 + x)^n &= \sum\limits_{i = 0}^{n}\binom{n}{i}x^i \\
\int\limits(1 + x)^ndx &= \sum\limits_{i = 0}^{n}\binom{n}{i}\int\limits{x^idx} \\
\frac{(x + 1)^{n + 1}}{n + 1} + C_1 &= \sum\limits_{i = 0}^{n}\frac{1}{i + 1}\binom{n}{i} x^{i + 1} + C_2 \\
\frac{1}{n + 1}\sum\limits_{i = 1}^{n + 1}\binom{n + 1}{i}x^i &= \sum\limits_{i = 0}^{n}\frac{1}{i + 1}\binom{n}{i} x^{i + 1} \\
&\xrightarrow{x = 1} \\
\sum\limits_{i = 0}^{n}\frac{1}{i + 1}\binom{n}{i} &= \frac{2^{n + 1} - 1}{n + 1}
\end{aligned}
\]
注意积分之后要把左边的常数项舍弃，否则最后推出来会少个 \(-1\)。
17.
16 的另解。
\[\sum\limits_{i = 0}^{n}\frac{1}{i + 1}\binom{n}{i} = \frac{1}{n + 1}\sum\limits_{i = 0}^{n}\binom{n + 1}{i + 1} = \frac{2^{n + 1} - 1}{n + 1}
\]
18.
\[\begin{aligned}
\sum\limits_{i = 0}^{n}(-1)^i\frac{1}{i + 1}\binom{n}{i} &= \frac{1}{n + 1}\sum\limits_{i = 0}^{n}(-1)^i\binom{n + 1}{i + 1} \\
&= \frac{1}{n + 1}\sum\limits_{i = 1}^{n + 1}(-1)^i\binom{n + 1}{i} \\
&= \frac{1}{n + 1}\left(\sum\limits_{i = 0}^{n + 1}(-1)^i\binom{n + 1}{i} - 1 \right) \\
&= -\frac{1}{n + 1}
\end{aligned}
\]
19.
\[2\binom{m}{2} + \binom{m}{1} = 2\left( \binom{m}{2} + \binom{m}{1} \right) - \binom{m}{1} = 2\binom{m + 1}{2} - m = m(m + 1) - m = m^2
\]
\[\sum\limits_{i = 1}^{n}i^2 = \sum\limits_{i = 1}^{n}2\binom{i}{2} + \binom{i}{1} = 2\sum\limits_{i = 1}^{n}\binom{i}{2} + \sum\limits_{i = 1}^{n}\binom{i}{1} = 2\binom{n + 1}{3} + \binom{n + 1}{2}
\]
20.
\[m^3 = \frac{am(m - 1)(m - 2)}{6} + \frac{bm(m - 1)}{2} + cm = \frac{a}{6}m^3 + \frac{b - a}{2}m^2 + \left( \frac{a}{3} - \frac{b}{2} + c \right)m
\]
\[\begin{aligned}
\begin{cases}
\frac{a}{6} = 1 \\
\frac{b - a}{2} = 0 \\
\frac{a}{3} - \frac{b}{2} + c = 0
\end{cases}
\end{aligned}
\]
解得 \(a = 6, b = 6, c = 1\)，\(\therefore m^3 = 6\binom{m}{3} + 6\binom{m}{2} + \binom{m}{1}\)。
\[\sum\limits_{i = 1}^{n}i^3 = 6\sum\limits_{i = 1}^{n}\binom{i}{3} + 6\sum\limits_{i = 1}^{n}\binom{i}{2} + \sum\limits_{i = 1}^{n}\binom{i}{1} = 6\binom{n + 1}{4} + 6\binom{n + 1}{3} + \binom{n + 1}{2}
\]
21.
\[(-1)^k\binom{k + r - 1}{k} = (-1)^k\frac{(k + r - 1)^{\underline{k}}}{(r - 1)!} = \frac{(1 - k - r)^{\overline{k}}}{(r - 1)!} = \frac{(1 - k - r + k - 1)^{\underline{k}}}{(r - 1)!} = \frac{(-r)^{\underline{k}}}{(r - 1)!} = \binom{-r}{k}
\]
22.
\[\binom{r}{m}\binom{m}{k} = \frac{r!}{m!(r - m)!}\frac{m!}{k!(m - k)!} = \frac{r!}{k!(r - m)!(m - k)!} = \frac{r!}{k!(r - k)!}\frac{(r - k)!}{(r - m)!(m - k)!} = \binom{r}{k}\binom{r - k}{m - k}
\]
