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),(
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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
![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),(](/uploads/image/z/400247-71-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca1%3D3%2Ca2%3D5%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3Sn%2BS%28n-2%29%3D2S%28n-1%29%2B2%5E+n-1%28n%E2%89%A53%29%2C%EF%BC%88)
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]
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]
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),(
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