已知数列｛an｝的首项为a1=2,前n项和为.Sn,且满足（an-1）n∧2＋n-Sn=0.(1已知数列｛an｝的首项为a1=2,_数学解答

已知数列｛an｝的首项为a1=2,前n项和为.Sn,且满足（an-1）n∧2＋n-Sn=0.(1
已知数列｛an｝的首项为a1=2,前n项和为.Sn,且满足（an-1）n∧2＋n-Sn=0.
(1)证明数列｛((n+1)/n )×Sn}是等差数列,并求数列｛an｝的通项公式.
(2)设bn=an/(n∧2+n+2),记数列bn的前n项和为Tn,证明Tn＜1

人气：216 ℃　时间：2020-05-17 01:53:02

解答

n=1时,S1=a1=2n≥2时,(an -1)n^2+n-Sn=0[Sn-S(n-1)-1]n^2+n-Sn=0(n^2-1)Sn- n^2S(n-1)=n^2-n(n+1)(n-1)Sn -n^2S(n-1)=n(n-1)等式两边同除以n(n-1)[(n+1)/n]Sn -[n/(n-1)]S(n-1)=1,为定值(2/1)S1=2S1=2×2=4,数列{[(n...