正数数列{an}的前n项和为Sn,且2√Sn=an +1 ,求⑴数列{an}的通项公式...答:正数数列{an}的前n项和为Sn,且2√Sn=an +1 ,求⑴数列{an}的通项公式,⑵设bn= 正数数列{an}的前n项和为Sn,且2√Sn=an+1,求⑴数列{an}的通项公式,⑵设bn=1/an*a(n+1),数列{bn}的前n项的和为Bn,求证2Bn<1详细过程,谢谢... 正数数列{an}的前n项和为Sn,且2√Sn=an +1 ,求⑴数列...
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...答:=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...