已知数列an的前n项和sn满足2(sn)^2=2ansn-an (n大于等于2),且a1=2,求an和sn
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![已知数列an的前n项和sn满足2(sn)^2=2ansn-an (n大于等于2),且a1=2,求an和sn](/uploads/image/z/2763671-23-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8Csn%E6%BB%A1%E8%B6%B32%28sn%29%5E2%3D2ansn-an+%28n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%EF%BC%89%2C%E4%B8%94a1%3D2%2C%E6%B1%82an%E5%92%8Csn)
已知数列an的前n项和sn满足2(sn)^2=2ansn-an (n大于等于2),且a1=2,求an和sn
已知数列an的前n项和sn满足2(sn)^2=2ansn-an (n大于等于2),且a1=2,求an和sn
已知数列an的前n项和sn满足2(sn)^2=2ansn-an (n大于等于2),且a1=2,求an和sn
2Sn^2=2anSn-an
an=Sn-S(n-1)
2Sn^2=2[Sn-S(n-1)]Sn-Sn+S(n-1)
=2Sn^2-2SnS(n-1)-Sn+S(n-1)
即2SnS(n-1)+Sn=S(n-1)
2SnS(n-1)+Sn=S(n-1) Sn=x,S(n-1)=y
2xy+x=y x=y/(2y+1)
1/x=2+1/y,即1/Sn=2+1/S(n-1)
所以1/Sn是以1/2为首项,2为差的等差数列
所以1/Sn=1/2+2(n-1)=(4n-3)/2 Sn=2/(4n-3)
1/Sn=2+1/S(n-1)=2+1/[Sn-an]
(4n-3)/2=2+1/[2/(4n-3)-an](移项,倒数,化简)
an=-8/[(4n-7)(4n-3)] n≥2
即an=-8/[(4n-7)(4n-3)] n≥2,Sn=2/(4n-3)
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