二项式反演
引入
当我在网上查找关于二项式反演的博客时， 总是只看到两个公式， 一个原式， 一个推论(1)， 所以这篇博客主要是引出推论(2)， 并且将其证明。
前两个式子
原式:
\[{\large f(n) = \sum \limits_{i = 0}^{n} \binom{n}{i}g(i) \Leftrightarrow g(n) = \sum \limits_{i = 0}^{n}(-1)^{n - i}\binom{n}{i}f(i)}
\]
推论(1)：
\[{\large f(n) = \sum \limits_{i = m}^{n} \binom{n}{i}g(i) \Leftrightarrow g(n) = \sum \limits_{i = m}^{n}(-1)^{n - i}\binom{n}{i}f(i)}
\]
这两个式子网上很多人证明， 这里也就不证了。
最关键的， 还是最后一个式子。
最后一个式子
推论(2):
\[{\large f(n) = \sum \limits_{i = n}^{m} \binom{i}{n}g(i) \Leftrightarrow g(n) = \sum \limits_{i = n}^{m}(-1)^{i - n}\binom{i}{n}f(i)}
\]
证明：
\[{\large f(n) = \sum \limits_{i = n}^{m} \binom{i}{n}g(i) \Leftrightarrow g(n) = \sum \limits_{i = n}^{m}(-1)^{i - n}\binom{i}{n}f(i)}
\]
若原式正确， 则有
\[{\large g(n) = \sum \limits_{i = n}^{m}(-1)^{i - n}\binom{i}{n}f(i)}\large\Leftrightarrow {\large g(n) = \sum \limits_{i = n}^{m}(-1)^{i - n}\binom{i}{n}\sum \limits_{j = i}^{m} \binom{j}{i}g(j)}
\]
\[\large\therefore {\large g(n) = \sum \limits_{i = n}^{m}(-1)^{i - n}\binom{i}{n}\sum \limits_{j = i}^{m} \binom{j}{i}g(j)}
\]
\[{\large = \sum \limits_{i = n}^{m} \sum\limits_{j = i}^{m} (-1)^{i - n}\binom{i}{n}\binom{j}{i}g(j)}
\]
\[{\large = \sum \limits_{j = n}^{m} \sum\limits_{i = n}^{j} (-1)^{i - n}\binom{i}{n}\binom{j}{i}g(j)}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j) \sum\limits_{i = n}^{j} (-1)^{i - n}\binom{i}{n}\binom{j}{i}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j) \sum\limits_{i = n}^{j} (-1)^{i - n}\binom{i}{n}\binom{j}{i}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j) \sum\limits_{i = n}^{j} (-1)^{i - n}\frac{i!}{n!(i-n)!}\frac{j!}{i!(j-i)!}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j) \sum\limits_{i = n}^{j} (-1)^{i - n}\frac{1}{n!(i-n)!}\frac{j!}{(j-i)!}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j) \sum\limits_{i = n}^{j} (-1)^{i - n}\frac{(j-n)!}{n!(i-n)!}\frac{j!}{(j-i)!(j-n)!}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j) \sum\limits_{i = n}^{j} (-1)^{i - n}\frac{(j-n)!}{(j-i)!(i-n)!}\frac{j!}{n!(j-n)!}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j) \sum\limits_{i = n}^{j} (-1)^{i - n}\binom{j - n}{i - n}\binom{j}{n}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j)\binom{j}{n} \sum\limits_{i = n}^{j} (-1)^{i - n}\binom{j - n}{i - n}}
\]
\[{\large = \sum \limits_{j = n}^{m} g(j)\binom{j}{n} \sum\limits_{i = 0}^{j - n} (-1)^{i}\binom{j - n}{i}}
\]
\[{\large 设k=j - n}
\]
\[{\large g(n) = \sum \limits_{j = n}^{m} g(j)\binom{j}{n} \sum\limits_{i = 0}^{k} (-1)^{i}\binom{k}{i}}
\]
\[{\large 又\because \sum\limits_{i = 0}^{k} (-1)^{i}\binom{k}{i} = }
{\large
\begin{aligned}
\left \{
\begin{aligned}
0, k\neq0,\\
1, k=0.
\end{aligned}
\right.
\end{aligned}
}
\]
\[{\large \therefore 当且仅当k = 0, 即j = n时\sum\limits_{i = 0}^{k} (-1)^{i}\binom{k}{i} = 1}
\]
\[{\large \therefore g(n) = g(n)\binom{n}{n} = g(n)}
\]
\[{\large \therefore 原命题显然成立}
\]
至此， 二项式反演的三个式子都弄完了， 至于第三个式子的应用。。。
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