令x=x,y=1:
for:f(x+y)=f(x)f(y)
so:f(x+1)=f(x)f(1)=f(x)/2
so:f(n)=f(1)/[2^(n-1)]=1/(2^n)
a1+a2+a3+.+an=1/2+1/4+…+1/(2^n)=(2^(n+1)-1)/(2^n)=2-1/(2^n)