已知正数数列an的前n项和为sn,满足sn^2=a1^3+.an^3.（1）求证an为等差数列,并求出通项公式（2）设bn=（1-1/an)^2-a(1-1/an）,若bn+1>bn对任意n属于N*恒成立,求实数a的取值范围
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习题大全
1Sn^2=a1^3a2^3an^3Sn1^2=a1^3a2^3an1^3an^3=Sn^2Sn1^2=SnSn1SnSn1=anSnSn1an0an^2=SnSn1n2an1^2=Sn1Sn2n2an2an12=SnSn2=anan1anan1=1n3S1^2=a1^2=a1^3a10a1=1S2^2=a1a2^2=a1^3a2^31a2^2=1a2^3a2^3a2^22a2=0a20a2=2anan1=1n2anan=n2bn=11n^2a11n^2=1n^2a2n1abn1bn=1n11n1n11na2=[1nn1][1n11na2]gt;01n11na2...
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