试解津京扩展的贫论员的问题

S(1) = a, S(2) = b, S(n+1)=n(S(n) + S(n-1)).
Find S(n) and limit(S(n)/n!).

Solution:
For n >=3,
  S(n)   = nS(n-1)       + (n-1)S(n-2) - S(n-1)
 -S(n-1) = -(n-1)S(n-2)  - (n-2)S(n-3) + s(n-2)
  S(n-2) = (n-2)S(n-3)   + (n-3)S(n-4) + S(n-3)
  ......
(-1)^(n-3)*S(3) = (-1)^(n-3)(3S(2) + 2S(1) - S(2)).
Sum up and let c = b - 2a

  S(n) = nS(n-1) + (-1)^n*c                         (1)

Claim: S(n) = n!( b/2 - c/3! + c/4! ... + (-1)^n c/n!).

n = 3, S(3) = 3!( b/2 - c/3! ) = 3b - (b - 2a) = 2(S(2) + S(1)).
From k to k+1 is a straight forward result from (1).

Now, T(n) = b/2 - c/3! + c/4! ... + (-1)^n c/n!
          = (b-c)/2 + c(1 - 1/1! + 1/2! + ... + (-1)^n /n! ).

So limit T(n) = (b-c)/2 + c/e = a + (b - 2a)/e.

所有跟帖： 

• 有趣！能否进一步推广系数n， S(n+1)=n1*S(n) + n2*S(n-1)。 -皆兄弟也- ♂ (0 bytes) () 10/15/2011 postreply 01:19:10

• 狗来一个结果： -wxcfan123- ♂ (186 bytes) () 10/15/2011 postreply 08:32:52

• 从结果可看出解法。但我的结果有点不一样。 -wxcfan123- ♂ (501 bytes) () 10/15/2011 postreply 13:27:01

• 请猜下面我的题,再给解法. -jinjing- ♀ (0 bytes) () 10/16/2011 postreply 07:21:27

• 没人答我的题,难过哦,老小老小,一点不假. -jinjing- ♀ (44 bytes) () 10/16/2011 postreply 07:53:04

• 题解不出来,陈露文家一点趣事...倒很想听 -皆兄弟也- ♂ (0 bytes) () 10/16/2011 postreply 22:13:54

• 题是笑话,陈家的事... -jinjing- ♀ (263 bytes) () 10/17/2011 postreply 14:12:43