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
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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
(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)
∴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)
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