表达
若有(f(n)=sum_{i=0}^ninom{n}{i}g(i))
则有(g(n)=sum_{i=0}^n(-1)^{n-i}inom{n}{i}f(i))
证明
(egin{aligned}g(n)&=sum_{i=0}^n(-1)^{n-i}inom{n}{i}f(i)\
&=sum_{i=0}^n(-1)^{n-i}inom{n}{i}sum_{j=0}^iinom{i}{j}g(j)\
&=sum_{i=0}^nsum_{j=0}^i(-1)^{n-i}inom{n}{i}inom{i}{j}g(j)\
&=sum_{j=0}^nsum_{i=j}^n(-1)^{n-i}inom{n}{i}inom{i}{j}g(j)\
end{aligned})
在组合数中，有这么一个式子
(egin{aligned}inom{i}{j}inom{j}{k} &= frac{i!}{j!(i - j)!}frac{j!}{k!(j - k)!} \
&=dfrac {i!}{k!left( i-k
ight) !}dfrac {left( i-k
ight) !}{left( i-j
ight) !left( j-k
ight) !}\
&=dfrac {i!}{k!left( i-k
ight) !}dfrac {left( i-k
ight) !}{left( left( i-k
ight) -left( j-k
ight)
ight) !left( j-k
ight) !}\
&=egin{pmatrix} i \ k end{pmatrix}egin{pmatrix} i-k \ j-k end{pmatrix}end{aligned})
即
(egin{aligned}inom{i}{j}inom{j}{k}=egin{pmatrix} i \ k end{pmatrix}egin{pmatrix} i-k \ j-k end{pmatrix}end{aligned})
所以
(egin{aligned}原式&=sum_{j=0}^nsum_{i=j}^n(-1)^{n-i}inom{n}{j}inom{n-j}{i-j}g(j)\
&=sum_{j=0}^nsum ^{n-j}_{i=0}left( -1
ight) ^{n-i-j}egin{pmatrix} n \ j end{pmatrix}egin{pmatrix} n-j \ i end{pmatrix}g(j)\
&=sum_{j=0}^nleft(1-1
ight)^{n-j}egin{pmatrix} n \ j end{pmatrix}gleft(j
ight)
end{aligned})
此时要求(n != j)，此时(原式=0)
当(n=j)时，(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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