Let me try

First we introduce new variable to normalize the old variables. Let
A_n=a_n– (n+1)/3
B_n=b_n– (n+1)/3
C_n=c_n– (n+1)/3.
Then A_n+B_n+C_n=0 and they satisfy the same recursion as a_n, b_n and c_n. We need to show that |A_n|=1.

Let S_n=A_n*A_n+B_n*B_n+C_n*C_n.Using the recursion relation, squaring on both side, addin up the three equations and using the relation that A_i=-(B_i+C_i) for each i, we have
S_n=S_{n-1}(1-1/n+1/n*n)
Now noticing that A_2=0 B_2=1/2 C_2=-1/2 hence S_2=1/2, we have
S_n = 2/n+2
Now with loss of generality, assuming A_n>0 and B_n, C_n = (1+1/2)A_n*A_n, or A_n*A_n

所有跟帖： 

• 回複：Let me try -藍糊塗- ♂ (40 bytes) () 01/15/2010 postreply 06:14:28

• 回複：回複：Let me try -innercool- ♀ (593 bytes) () 01/15/2010 postreply 08:33:55

• 回複：回複：回複：Let me try -藍糊塗- ♂ (0 bytes) () 01/15/2010 postreply 08:39:07

• 回複：回複：回複：Let me try -藍糊塗- ♂ (5 bytes) () 01/15/2010 postreply 08:41:14

• 回複：回複：回複：Let me try -jinjing- ♀ (73 bytes) () 01/15/2010 postreply 11:39:57

• master jinjing不得了。數學頂瓜瓜，提琴也會拉：） -戲雨飛鷹- ♀ (0 bytes) () 01/15/2010 postreply 18:42:08