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f(n)=\sum_{i=0}^n\binom{n}{i}g(i)
f(n)=∑i=0n(in)g(i)
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g(n)=\sum_{i=0}^n(-1)^{n-i}\binom{n}{i}f(i)
g(n)=∑i=0n(−1)n−i(in)f(i)
证明
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\begin{aligned}g(n)&=\sum_{i=0}^n(-1)^{n-i}\binom{n}{i}f(i)\\ &=\sum_{i=0}^n(-1)^{n-i}\binom{n}{i}\sum_{j=0}^i\binom{i}{j}g(j)\\ &=\sum_{i=0}^n\sum_{j=0}^i(-1)^{n-i}\binom{n}{i}\binom{i}{j}g(j)\\ &=\sum_{j=0}^n\sum_{i=j}^n(-1)^{n-i}\binom{n}{i}\binom{i}{j}g(j)\\ \end{aligned}
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∑nj=0∑i(−1)n−i(in)(ji)g(j)=j=0∑ni=j∑n(−1)n−i(in)(ji)g(j)
在组合数中，有这么一个式子
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\begin{aligned}\binom{i}{j}\binom{j}{k} &= \frac{i!}{j!(i - j)!}\frac{j!}{k!(j - k)!} \\ &=\dfrac {i!}{k!\left( i-k\right) !}\dfrac {\left( i-k\right) !}{\left( i-j\right) !\left( j-k\right) !}\\ &=\dfrac {i!}{k!\left( i-k\right) !}\dfrac {\left( i-k\right) !}{\left( \left( i-k\right) -\left( j-k\right) \right) !\left( j-k\right) !}\\ &=\begin{pmatrix} i \\ k \end{pmatrix}\begin{pmatrix} i-k \\ j-k \end{pmatrix}\end{aligned}
(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
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\begin{aligned}\binom{i}{j}\binom{j}{k}=\begin{pmatrix} i \\ k \end{pmatrix}\begin{pmatrix} i-k \\ j-k \end{pmatrix}\end{aligned}
(ji)(kj)=(ik)(i−kj−k)
所以
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\begin{aligned}原式&=\sum_{j=0}^n\sum_{i=j}^n(-1)^{n-i}\binom{n}{j}\binom{n-j}{i-j}g(j)\\ &=\sum_{j=0}^n\sum ^{n-j}_{i=0}\left( -1\right) ^{n-i-j}\begin{pmatrix} n \\ j \end{pmatrix}\begin{pmatrix} n-j \\ i \end{pmatrix}g(j)\\ &=\sum_{j=0}^n\left(1-1\right)^{n-j}\begin{pmatrix} n \\ j \end{pmatrix}g\left(j\right) \end{aligned}
原式=j=0∑ni=j∑n(−1)n−i(jn)(i−jn−j)g(j)=j=0∑ni=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
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n!=j，此时
原
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原式=0
当
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原
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原式=g(n)
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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