Question

已知(2x-1)的6次方=ax6次+bx5次+cx4次+dx3次+ex2+fx+g 求a+b+c+d+e+f+g=?求a

Answer 1

x=1则a+b+c+d+e+f+g=(1-3)^6=64x=-1则a-b+c-d+e-f+g=(-1-3)^6=4096-a+b-c+d-e+f-g=-(a-b+c-d+e-f+g)=-4096a+b+c+d+e+f+g=64a-b+c-d+e-f+g=4096相减2(b+d+f）=-4032b+d+f=-2016-a-c-e-g=(-a+b-c+d-e+f-g)-(b+d+f)=-20...