第1个回答 2019-02-08
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)