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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.
人气:206 ℃ 时间:2020-03-25 21:04:04
解答
(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=-242②
①②联立解得a+c+e=122.
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