若(3x+1)^5=ax^5+bx^5+cx^3+dx^2+ex+f,则a-b+c-d+e-f的值是

问题描述:

若(3x+1)^5=ax^5+bx^5+cx^3+dx^2+ex+f,则a-b+c-d+e-f的值是

令x=-1
则(-3+1)^5=-a+b-c+d-e+f=-2^5
所以a-b+c-d+e-f=2^5=32

令x=-1
a-b+c-d+e-f的值是32

令x=-1
那么a-b+c-d+e-f=-(ax^5+bx^5+cx^3+dx^2+ex+f)=-(3x+1)^5=32