(Ⅰ)n
2时,f(n)=S n﹣S n﹣1=2n+1.
n=1时,f(1)=S 1=3,适合上式,
∴f(n)=Sn﹣Sn﹣1=2n+1.(n∈N*).
(Ⅱ)a 1=f(1)=3,a n+1=2a n+1,(n∈N*).即a n+1+1=2(a n+1).
∴数列{an+1}是首项为4,公比为2的等比数列.a n+1=(a 1+1)2n+1=2n+1.
a n=2n+1﹣1,(n∈N*).
T n=2 2+2 3+2 4+…+2 n+1﹣n=2 n+2﹣4﹣n.
(Ⅰ)n
2时,f(n)=S n﹣S n﹣1=2n+1.
n=1时,f(1)=S 1=3,适合上式,
∴f(n)=Sn﹣Sn﹣1=2n+1.(n∈N*).
(Ⅱ)a 1=f(1)=3,a n+1=2a n+1,(n∈N*).即a n+1+1=2(a n+1).
∴数列{an+1}是首项为4,公比为2的等比数列.a n+1=(a 1+1)2n+1=2n+1.
a n=2n+1﹣1,(n∈N*).
T n=2 2+2 3+2 4+…+2 n+1﹣n=2 n+2﹣4﹣n.