[f(n)=sum_{i=0}^n{nchoose i}g(i) iff g(n)=sum_{i=0}^n(-1)^{(n-i)}{nchoose i} f(i)
]
证明如下：
[g(n)=sum_{i=0}^n(-1)^{n-i}{n choose i} sum_{j=0}^i {ichoose j}g(j)\
g(n)=sum_{i=0}^n sum_{j=0}^i (-1)^{n-i}{n choose i} {ichoose j}g(j)\
g(n)=sum_{j=0}^n sum_{i=j}^n (-1)^{n-i}{n choose i} {ichoose j}g(j)\
]
[{i choose j}{j choose k}=frac{i!j!}{j!(i-j)!k!(j-k)!}\
=frac{i!(i-k)!}{(i-j)!(i-k)!k!(j-k)!}\
=frac{i!}{k!(i-k)!}frac{(i-k)!}{(i-j)!(j-k)!}\
=frac{i!}{k!(i-k)!}frac{(i-k)!}{(j-k)!((i-k)-(j-k))!}\
={i choose k}{i-k choose j-k}
]
[g(n)=sum_{j=0}^n sum_{i=j}^n (-1)^{n-i}{n choose j} {n-jchoose i-j}g(j)\
g(n)=sum_{j=0}^n sum_{i=0}^{n-j} (-1)^{n-i-j}{n-jchoose i} {n choose j} g(j)\
]
[sum_{i=0}^{n-j} (-1)^{n-i-j}{n-jchoose i}1^{i}\
(1-1)^{n-j}
]
[g(n)=sum_{j=0}^n (1-1)^{n-j}{n choose j} g(j)\
g(n)=g(n)\
]