一道数学题已知ax+by=7,ax^2+by^2=49,ax^3+by^3=133,ax^4+by^4=406.求2008（x+y）+2008xy+（a+b）/21

Question

一道数学题
已知ax+by=7,ax^2+by^2=49,ax^3+by^3=133,ax^4+by^4=406.求2008（x+y）+2008xy+（a+b）/21

Answer 1

设
x+y = m
xy = n
a + b = k
(x+y)(ax+by) = ax^2+by^2+(a+b)xy =>7m = 49+kn (1)
(x+y)(ax^2+by^2) = ax^3+by^3+(ax+by)xy => 49m = 133 + 7n (2)
(x+y)(ax^3+by^3) = ax^4+by^4+(ax^2+by^2)xy => 133m = 406 + 49n (3)
由(2)(3)得
m = 5/2
n = -3/2
代入(1)得k = 21
所以2008(x+y) + 2008xy - (a+b) /21= 2008* 5/2 - 2008*(3/2) - 21/21 =2008-1=2007