Question

已知f(x)=a1x+a2x^2+…+anx^n,则a1,a2,…an组成等差数列,f（1）=n^2,an=2n-1

Answer 1

f(1)=a1+……+an=n^2

a1=1,a2=3,……,an=2n-1

f(1/2)=S=1*(1/2)+3*(1/2)^2+……+(2n-1)*(1/2)^n

2S=1+3*(1/2)+……+(2n-1)*(1/2)^(n-1)

所以f(1/2)=S=2S-S=1+2*(1/2)+2*(1/2)^2+……+2*(1/2)^(n-1)-(2n-1)*(1/2)^n

=1+2*(1/2)*[1-(1/2)^(n-1)]/(1-1/2)-(2n-1)*(1/2)^n

=3-(1/2)^(n-1)-(2n-1)*(1/2)^n