已知数列{an}中,a1=3,a2=5,其前n项和Sn满足Sn+S(n-2)=2S(n-1)+2^ n-1(n≥3),（2）令bn=1/an*a(n+1),Tn是
问题描述：
已知数列{an}中,a1=3,a2=5,其前n项和Sn满足Sn+S(n-2)=2S(n-1)+2^ n-1(n≥3),（2）令bn=1/an*a(n+1),Tn是
答
Sn+S(n-2)=2S(n-1)+2^ (n-1)
Sn-S(n-1)-[S(n-1)-S(n-2)]=2^(n-1)
an-a(n-1)=2^(n-1)
a(n-1)-a(n-2)=2^(n-2)
·····
a3-a2=2^2
an-a2=2^n-4(n>=3)
an=2^n+1
对于a1=3,a2=5都满足
故an=2^n+1
2;bn=1/an*a(n+1)
=1/(2^n+1)[2^(n+1)+1]
bnf(x)=2^(n-1)/{(2^n+1)[2^(n+1)+1]}
=1/2{1/(2^n+1)-1/[2^(n+1)+1]}
所以 Tn=b1f(1)+b2f(2)+...bnf(n)
=1/2{1/3-1/5+1/5-`````-1/[2^(n+1)+1]
=1/6-1/[2^(n+1)+1]