Question

正项数列{an}满足an²–（2n－1）an－2n＝0求数列an的通项公式 令bn＝1/(n+1)an求数列bn的前n项和

Answer 1

an²–（2n－1）an－2n＝0
(an +1)(an-2n)=0
an=-1 舍去
an=2n
bn=1/(n+1)*2n
Sn= 1/4 + 1/12 +.... +1/(n+1)*2n
=1/2(1-1/2 +1/2-1/3+...+1/n-1/(n+1)
=1/2 *(1-1/n+1)
=n/(2n+2)

追问

谢谢

Answer 2

(1)
(an)^2-(2n-1)an-2n=0
(an-2n)(an+1)=0
an=2n
(2)
bn=1/[(n+1).an]
= (1/2)[ 1/n -1/(n+1)]
Tn = b1+b2+..+bn
= (1/2) 1- 1/(n+1))