已知数列an中,a1=1,当n≥2时,其前n项和Sn满足Sn^2=an(Sn-1/2)
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已知数列an中,a1=1,当n≥2时,其前n项和Sn满足Sn^2=an(Sn-1/2)
(1)求Sn的表达式
(2)设bn=Sn/2n+1,求bn的前n项和Tn
(1)求Sn的表达式
(2)设bn=Sn/2n+1,求bn的前n项和Tn
![已知数列an中,a1=1,当n≥2时,其前n项和Sn满足Sn^2=an(Sn-1/2)](/uploads/image/z/645770-2-0.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D1%2C%E5%BD%93n%E2%89%A52%E6%97%B6%2C%E5%85%B6%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3Sn%5E2%3Dan%28Sn-1%2F2%29)
(Sn)²=[Sn-S(n-1)](Sn-1/2)
(Sn)²=(Sn)²-Sn/2-SnS(n-1)+S(n-1)/2
Sn+2SnS(n-1)-S(n-1)=0
S(n-1)-Sn=2SnS(n-1)
两边除以SnS(n-1)
1/Sn-1/S(n-1)=2
1/Sn等差,d=2
S1=a1=1
1/Sn=1/S1+2(n-1)=2n-1
Sn=1/(2n-1)
bn=1//[(2n-1)(2n+1)]
=1/2*2[(2n-1)(2n+1)]
=1/2*[(2n+1)-(2n+1)]/[(2n-1)(2n+1)]
=1/2*{(2n+1)/[(2n-1)(2n+1)]-(2n+1)/[(2n-1)(2n+1)]}
=1/2*[1/[(2n-1)-1/(2n+1)]
所以Tn=1/2*(1-1/3+1/3-1/5+1/5-1/7+……+1/[(2n-1)-1/(2n+1)]
=1/2*(1-1/(2n+1)]
=n/(2n+1)
(Sn)²=(Sn)²-Sn/2-SnS(n-1)+S(n-1)/2
Sn+2SnS(n-1)-S(n-1)=0
S(n-1)-Sn=2SnS(n-1)
两边除以SnS(n-1)
1/Sn-1/S(n-1)=2
1/Sn等差,d=2
S1=a1=1
1/Sn=1/S1+2(n-1)=2n-1
Sn=1/(2n-1)
bn=1//[(2n-1)(2n+1)]
=1/2*2[(2n-1)(2n+1)]
=1/2*[(2n+1)-(2n+1)]/[(2n-1)(2n+1)]
=1/2*{(2n+1)/[(2n-1)(2n+1)]-(2n+1)/[(2n-1)(2n+1)]}
=1/2*[1/[(2n-1)-1/(2n+1)]
所以Tn=1/2*(1-1/3+1/3-1/5+1/5-1/7+……+1/[(2n-1)-1/(2n+1)]
=1/2*(1-1/(2n+1)]
=n/(2n+1)
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