I got n-1 independent equations.why do you use b2+c3b3+...+cnbn?

（b2+c3b3+...+cnbn）a1 +（b2+c3b3+...+cnbn）b1x = a2+b2x

a2+b2x = c3a3+c3b3x

c3a3+c3b3x = c4a4+c4b4x
...

c(n-1)a(n-1) + c(n-1)b(n-1) x = cnan+cnbnx

unknowns are x, c3,...,c(n-1), cn.

所有跟帖： 

• n=2,your first Q is wrong. In fact t, you can choice any (n-1) -jinjing- ♀ (113 bytes) () 09/14/2010 postreply 10:41:32

• I have realized that it is wrong. -皆兄弟也- ♂ (0 bytes) () 09/14/2010 postreply 10:53:35

• The sum of all factors should be constant(no x). -jinjing- ♀ (150 bytes) () 09/15/2010 postreply 10:20:17