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
(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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