问题
填空题
若(2x+1)100=a0+a1(x-1)+a2(x-1)2+…+a100(x-1)100,则a1+a3+a5+…+a99=______.
答案
令x=2得
5100=a0+a1+a2+…+a100,
令x=0得
1=a0-a1+a2-a3…+a100,
两式相减得
=a1+a3+a5+…+a995100-1 2
故答案为
.5100-1 2
若(2x+1)100=a0+a1(x-1)+a2(x-1)2+…+a100(x-1)100,则a1+a3+a5+…+a99=______.
令x=2得
5100=a0+a1+a2+…+a100,
令x=0得
1=a0-a1+a2-a3…+a100,
两式相减得
=a1+a3+a5+…+a995100-1 2
故答案为
.5100-1 2