Question

如何计算n!=1!+2!+3!+…!

Answer 1

解: D1=a+b, D2=a^2+ab+b^2.
n>2时,将Dn按第一列展开
得 Dn=(a+b)Dn-1 - abDn-2 (1)
所以 Dn-aDn-1 = b(Dn-1-aDn-2)
= b^2(Dn-2-aDn-3) --迭代
= ...
= b^(n-2)(D2-aD1) = b^(n-2)b^2
= b^n. (2)
由(1)式同理可得
Dn-bDn-1 = a(Dn-1-bDn-2) = a^n (3)

若 a=b, 由(3)
Dn=aDn-1+a^n
= a(aDn-2+a^(n-1) +a^n = a^2Dn-2 + 2a^n
= ...
= a^(n-1)D1+(n-1)a^n
= (n+1)a^n. (4)
若 a≠b, 由 a(3)-b(2) 得
(a-b)Dn = a^(n+1) - b^(n+1)
所以 Dn = [a^(n+1)-b^(n+1)]/(a-b) (5)

(4)(5)合并为 ∑[k:0,n]a^kb^(n-k)