an=a1*q^(n-1)
1：a3^2=2a2a5,a1^2*q^4=2a1*q*a1*q^4
得q=1/2,
a1+2*a1*q=1,得
a1=1/2
故an=1/2^n
an=2^(-n)
2:bn=log2(a1*a2*a3*.*an)
bn=log2[2^(-1)*2(-2)*2^(-3)****2^(-n)]
bn=-1+(-2)+(-3)+.+(-n)
易证bn=-n（n+1）/2
数列{bn分之一}的通项公式cn=-2/[n*(n+1)]
c1=-1,cn=-2[1/n-1/(n+1)]
前n项和sn=-2[1-1/2+1/2-1/3+1/3-1/4+.-1/n+1/n-1/(n+1)]
sn=-2[1-1/(n+1)]
sn=-2n/(n+1)