回复：回复：Let me try

By the recursion equations for A_n, B_n and C_n we have

A_n=A_{n-1}+C_{n-1}/n
B_n=B_{n-1}+A_{n-1}/n
C_n=C_{n-1}+B_{n-1}/n.

Now squaring both sides and adding these three equations together:
S_n=(1+1/n*n)S_{n-1}+2/n(A_{n-1}*B_{n-1}+B_{n-1}*C_{n-1}+C_{n-1}*A_{n-1}).

Since A_{i}+B_{i}+C_{i}=0 for every i,
A_i*B_i+B_i*C_i+C_i*A_i
=-(B_i+C_i)*(B_i+C_i)+B_i*C_i
=-(B_i*B_i+C_i*C_i+B_i*C_i)
=-1/2(2*B_i*B_i+2*C_i*C_i+2B_i*C_i)
=-1/2(A_i*A_i+B_i*B_i+C_i*C_i)
=-1/2*S_{i}.
Now letting i=n-1 gives the desired equation.

On the other hand,

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• 回复：回复：回复：Let me try -蓝糊涂- ♂ (0 bytes) () 01/15/2010 postreply 08:39:07

• 回复：回复：回复：Let me try -蓝糊涂- ♂ (5 bytes) () 01/15/2010 postreply 08:41:14

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

• master jinjing不得了。数学顶瓜瓜，提琴也会拉：） -戏雨飞鹰- ♀ (0 bytes) () 01/15/2010 postreply 18:42:08