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
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/26 06:15:06
![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](/uploads/image/z/1741153-49-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%AD%A3%E6%95%B0%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E6%BB%A1%E8%B6%B3sn%5E2%3Da1%5E3%2B.an%5E3.%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81an%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82%E5%87%BA%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%BEbn%3D%EF%BC%881-1%2Fan%29%5E2-a%281-1%2Fan%EF%BC%89%2C%E8%8B%A5bn%2B1%3Ebn%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E5%B1%9E%E4%BA%8EN%2A%E6%81%92%E6%88%90%E7%AB%8B%2C%E6%B1%82%E5%AE%9E%E6%95%B0a%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4)
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
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
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
1、∵Sn^2=a1^3+a2^3+…+an^3,
∴Sn-1^2=a1^3+a2^3+…+a(n-1)^3,
两式相减,得an^3=Sn^2-S(n-1)^2=(Sn-S(n-1)))(Sn+S(n-1)))=an(Sn+S(n-1)),
∵an>0,∴an^2=Sn+S(n-1)(n≥2),
∴a(n-1 )^2=S(n-1)+S(n-2()n≥2),
两式相减,得an2-an-12 =Sn-S(n-2)=an+a(n-1),
∴an-a(n-1)=1(n>3),
∵S1^2=a1^2=a1^3,且a1>0,∴a1=1,
S2^2=(a1+a2)^2=a1^3+a2^3,
∴(1+a2)^2=1+a2^3,∴a2^3-a2^2-2a2=0,
由a2>0,得a2=2,
∴an-a(n-1)=1,n≥2,
故数列{an}为等差数列,通项公式为an=n.
2、bn=(1-1/n)^2-a(1-1/n)^2=1/n^2+(a-2)/n+1-a,
b(n+1)-bn=(1/(n+1)-1/n)(1/(n+1)+1/n+a-2)=-[1/n(n+1)][1/(n+1)+1/n+a-2]>0
即1/(n+1)+1/n+a-2)