Two expressions are subtracted: a (n+1)-an = 2 [sn-s (n-1)] = 2an, then a(n+ 1)=3an.

To make the series {an} a geometric series, the common ratio of this series can only be 3.

When n= 1, a2 = 2s1+1= 2a1+1= 2t+1,while a1= t.

So a2=3a 1=3t=2t+ 1, so t= 1.

(2)a 1=t= 1, so the sequence {an} is a geometric series with 1 as the geometric series and 3 as the common ratio.

Then a (n+ 1) = 1 * 3 n = 3 n.

So bn = log3 [a (n+1)] = n.

Then1/[bn * b (n+1)] =1[n (n+1)] =1/(n+/kloc.

So TN =1-1/2+1/2-1/3+…+1/n-1(n+1).

= 1- 1/(n+ 1)

=n/(n+ 1)

So t 2011= 2011/2012.