1.
证：
n=1时,1/S1=1/a1=1/(2/9)=9/2
n≥2时,
1/Sn-1/S(n-1)=[S(n-1)-Sn]/[SnS(n-1)]=-an/[SnS(n-1)]
an=SnS(n-1)代入,1/Sn-1/S(n-1)=-an/an=-1,为定值.
数列{1/Sn}是以9/2为首项,-1为公差的等差数列.
2.
1/Sn=9/2+(-1)(n-1)=11/2 -n
Tn=1/S1+1/S2+...+1/Sn
=11n/2 -(1+2+...+n)
=11n/2 -n(n+1)/2
=(n/2)(11-n-1)
=n(10-n)/2