problems on sequences.

Ha!
Let f(n) = a(n) /n for n=1,2,3,..
Then
f(n+1) = 1/ [2-f(n)] which can be solved analytically

2011-07-03 19:33:59 補充：
Ref: http://hk.knowledge.yahoo.com/question/question?qid=7010071501617

2011-07-03 19:39:04 補充：
Let g(n) = 2-f(n)
then
2 - g(n+1) = 1/g(n)
g(n+1) = 2 - 1/g(n) for n=1,2,...

Let h(n) = g(1)g(2)...g(n) n=1,2,...
then

multiply the above by g(n)g(n-1)...g(1),
h(n+1) = 2h(n) - h(n-1) n=2,3
==> ordinary diff. equation

2011-07-03 19:39:51 補充：
Very complicated

2011-07-04 15:42:27 補充：
As follows:

圖片參考：http://imgcld.yimg.com/8/n/HA00140412/o/701106220053713873462342.jpg

1. There are something wrong in (2).
2. Can it be solved analytically?

Given a(1)=a，a(n+1)= n(n+1)/[2n－a(n)] for n=2，3，4，...，find
(1) a(n)=?
Sol
a(2)=1*2/(2－a)=2/(2－a)
a(3)=2*3/[4－2/(2－a)]=6(2－a)/[8－4a－2]=6(2－a)/(6－4a)=3(2－a)/(3－2a)
a(4)=3*4/[6－3(2－a)/(3－2a)]=12(3－2a)/[18－12a－6+ 3a ]=12(3－2a)/(12－9a)
=4(3－2a)/(4－3a)
a(5)=4*5/[8－4(3－2a)/(4－3a)]=4*5*(4－3a)/[32－24a－12+ 8a ]
=40(4－3a)/(20－16a)=5(4－3a)/(5－4a)
……
a(n)=n[n－1－(n－2)a]/[n－(n－1)a]
(2) a(1)*a(2)*a(3)*...*a(n+1)/ n! = ?
Sol
a(1)*a(2)*a(3)*...*a(n+1)/ n!
=a*[2/(2－a)]*[3(2－a)/(3－2a)]*[4(3－2a)/(4－3a)]*…*
｛(n－1)[n－2－(n－3)a]/[n－1－(n－2)a]｝*｛n[n－1－(n－2)a]/[n－(n－1)a]｝/n!
=｛a*2*3*4*…*n/[n－(n－1)a]｝/n!
=a/[n－(n－1)a]
=a/(n－na+a)