08/01/26 20:21:54
>>148 (1) 再掲
0≦m≦n かつ m≦k≦n とする。
C[n,k] k(k-1)…(k-m+1) = C[n,k] {k!/(k-m)!} = {(n!)/(k!･(n-k)!)}{k!/(k-m)!} = {n!/(n-m)!} C[n-m,k-m],
より
∑[k=0,n] (-1)^k･C[n,k] k(k-1)…(k-m+1)
= ∑[k=m,n] (-1)^k･C[n,k] k(k-1)…(k-m+1)
= {n!/(n-m)!} Σ[k=m,n] (-1)^k･C[n-m,k-m]
= {n!/(n-m)!}(-1)^m Σ[k'=0,n-m] (-1)^k'･C[n-m,k']
= {n!/(n-m)!}(-1)^m･(1-1)^(n-m)
= (-1)^n･n!δ_(m,n),
よって 0≦m≦n のとき
∑[k=0,n] (-1)^k･C[n,k] k^m = (-1)^n･n!δ_(m,n),