设(2x-1)5=ax5+bx4+cx3+dx2+ex+f, 求:(1)f的值; (2)a+b+c+d+e+f的值; (3)a+c+e的值.
问题描述:
设(2x-1)5=ax5+bx4+cx3+dx2+ex+f,
求:(1)f的值;
(2)a+b+c+d+e+f的值;
(3)a+c+e的值.
答
(1)令x=0,ax5+bx4+cx3+dx2+ex+f=f=-1.(2)令x=1,ax5+bx4+cx3+dx2+ex+f=a+b+c+d+e+f=1,∴a+b+c+d+e=2 ①;(3)令x=-1,ax5+bx4+cx3+dx2+ex+f=-a+b-c+d-e+f=(-3)5=-243,∴-a+b-c+d-e=-242②①②联立解得a+c...