已知﹛an﹜的前n项和为Sn,若a1=1,Sn=nan-n(n-1),令bn=1/an*an+1,且数列的前项和为Tn

1.Sn=nan-n(n-1).1
所以S(n+1)=(n+1)a(n+1)-n(n+1).2
2式减1式：a(n+1)=(n+1)a(n+1)-nan-2n
所以na(n+1)-nan=2n即a(n+1)-an=2
所以an=2n-1
2.bn=1/[(2n-1)(2n+1)]
Tn=1/(1*3)+1/(3*5)+.+1/[(2n-1)(2n+1)]
=1/2(1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1))
=(1/2)*2n/(2n+1)
=n/(2n+1)
即λn/(2n+1)