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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
人气:267 ℃ 时间:2020-01-29 12:25:51
解答
(Sn)²=[Sn-S(n-1)](Sn-1/2)(Sn)²=(Sn)²-Sn/2-SnS(n-1)+S(n-1)/2Sn+2SnS(n-1)-S(n-1)=0S(n-1)-Sn=2SnS(n-1)两边除以SnS(n-1)1/Sn-1/S(n-1)=21/Sn等差,d=2S1=a1=11/Sn=1/S1+2(n-1)=2n-1Sn=1/(2n-1)bn=1...
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