a1a7=4,a6=8,
(a1)^2q^6=4 ==>a1q^3=2,
a1q^5=8
相除:q^2=4 ==>q=2
∴a1=1/4
an=1/4*2^(n-1)=2^(n-3)
f=a1x+a2x^2+a3x^3+...+a10x^10
f'(x)=a1+2a2x+3a3x^2+4a4x^3+...+10a10x^9
第n项bn=n*anx^(n-1)
x=1/2时,bn=n*2^(n-3)*(1/2)^(n-1)=n*2^(-2)=n/4
f'(1/2)=1/4+2/4+...+10/4=55/4