Question

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

Answer 1

(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)

Answer 2

2an=SnS(n-1)
an=Sn-S(n-1)
所以2(Sn-S(n-1))=Sn*S(n-1)
左右同除Sn*S(n-1)得到
2/S(n-1)-2/Sn=1
所以1/Sn-1/S(n-1)=-1/2
又S1=a1=3
所以｛1/Sn｝是首项为1/3，公差为-1/2的等差数列
所以1/Sn=-n/2+5/6
所以Sn=6/(5-3n)
S(n-1)=6/(8-3n)
因为2an=SnS(n-1)
所以an=18/[(5-3n)(8-3n)]

Answer 3

数项有n+1
偶数项是n

则奇数和=[a1+a(2n+1)](n+1)/2=290
偶数和=[a2+a(2n)]n/2=261
等差则a1+a(2n+1)=a2+a2n
所以相除有(n+1)/n=290/261=10/9
n=9
2n+1=19