也许更好的阅读体验

表达

若有 $f\left(n\right)={\sum }_{i=0}^n\left(\frac{n}{i}\right)g\left(i\right)$f(n)=∑i=0n​(in​)g(i)
则有 $g\left(n\right)={\sum }_{i=0}^n(-1{)}^{n-i}\left(\frac{n}{i}\right)f(i)$g(n)=∑i=0n​(−1)n−i(in​)f(i)

证明

$\begin{array}{cc}{\displaystyle g\left(n\right)}& {\displaystyle =\sum _{i=0}^n(-1{)}^{n-i}\left(\frac{n}{i}\right)f(i)}\\ {\displaystyle }& {\displaystyle =\sum _{i=0}^n(-1{)}^{n-i}\left(\frac{n}{i}\right)\sum _{j=0}^i\left(\frac{i}{j}\right)g(j)}\\ {\displaystyle }& {\displaystyle =\sum _{i=0}^n\sum _{j=0}^i(-1{)}^{n-i}\left(\frac{n}{i}\right)\left(\frac{i}{j}\right)g(j)}\\ {\displaystyle }& {\displaystyle =\sum _{j=0}^n\sum _{i=j}^n(-1{)}^{n-i}\left(\frac{n}{i}\right)\left(\frac{i}{j}\right)g(j)}\end{array}$g(n)​=i=0∑n​(−1)n−i(in​)f(i)=i=0∑n​(−1)n−i(in​)j=0∑i​(ji​)g(j)=i=0∑n​j=0∑i​(−1)n−i(in​)(ji​)g(j)=j=0∑n​i=j∑n​(−1)n−i(in​)(ji​)g(j)​
在组合数中，有这么一个式子
$\begin{array}{cc}{\displaystyle \left(\frac{i}{j}\right)\left(\frac{j}{k}\right)}& {\displaystyle =\frac{i!}{j!(i-j)!}\frac{j!}{k!(j-k)!}}\\ {\displaystyle }& {\displaystyle =\frac{i!}{k!(i-k)!}\frac{(i-k)!}{(i-j)!(j-k)!}}\\ {\displaystyle }& {\displaystyle =\frac{i!}{k!(i-k)!}\frac{(i-k)!}{((i-k)-(j-k))!(j-k)!}}\\ {\displaystyle }& {\displaystyle =\left(\begin{array}{c}i\\ k\end{array}\right)\left(\begin{array}{c}i-k\\ j-k\end{array}\right)}\end{array}$(ji​)(kj​)​=j!(i−j)!i!​k!(j−k)!j!​=k!(i−k)!i!​(i−j)!(j−k)!(i−k)!​=k!(i−k)!i!​((i−k)−(j−k))!(j−k)!(i−k)!​=(ik​)(i−kj−k​)​
即
$\begin{array}{c}{\displaystyle \left(\frac{i}{j}\right)\left(\frac{j}{k}\right)=\left(\begin{array}{c}i\\ k\end{array}\right)\left(\begin{array}{c}i-k\\ j-k\end{array}\right)}\end{array}$(ji​)(kj​)=(ik​)(i−kj−k​)​
所以
$\begin{array}{cc}{\displaystyle 原式}& {\displaystyle =\sum _{j=0}^n\sum _{i=j}^n(-1{)}^{n-i}\left(\frac{n}{j}\right)\left(\frac{n-j}{i-j}\right)g(j)}\\ {\displaystyle }& {\displaystyle =\sum _{j=0}^n\sum _{i=0}^{n-j}{(-1)}^{n-i-j}\left(\begin{array}{c}n\\ j\end{array}\right)\left(\begin{array}{c}n-j\\ i\end{array}\right)g\left(j\right)}\\ {\displaystyle }& {\displaystyle =\sum _{j=0}^n{(1-1)}^{n-j}\left(\begin{array}{c}n\\ j\end{array}\right)g\left(j\right)}\end{array}$原式​=j=0∑n​i=j∑n​(−1)n−i(jn​)(i−jn−j​)g(j)=j=0∑n​i=0∑n−j​(−1)n−i−j(nj​)(n−ji​)g(j)=j=0∑n​(1−1)n−j(nj​)g(j)​
此时要求 $n!=j$n!=j，此时 $原式=0$原式=0
当 $n=j$n=j时， $原式=g\left(n\right)$原式=g(n)

到此，原式得证

如有哪里讲得不是很明白或是有错误，欢迎指正
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