设等比为q,则a2=a1*q,a3=a1*q^2,a6=a1*q^5
则:2a1+3a2=2a1+3a1*q=1,(a1*q^2)^2=9*(a1*q)*(a1*q^5)
化简得:2a1+3a1*q=1①,q^4=9*q*q^5②
由②式得q^2=1/9,因为等比数列an的各项为正数,即q>0,所以q=1/3,代入①式得:a1=1/3
所以an的通项公式为an=a1*q^(n-1)=(1/3)*(1/3)^(n-1)=(1/3)^n (n≥1)
2.
bn=-1-2-....n=-n(n+1)/2
1/bn=-2/[n(n+1)]=-2[1/n-1/(n+1)]
Sn=-2[1-1/2+1/2-1/3+...+1/n-1/(n+1)]=-2[1-1/(n+1)]=-2n/(n+1)