题目

设数列{an}的前n项和为Sn,S2=5,a(n+1)=an+2n+1.
(1)求an的通项公式
(2)设Tn=1/a1+1/a2+...+1/an,求证nTn

提问时间：2021-03-12

答案

(1)
S2 =5
a1+a2= 5
a(n+1)=an+2n+1
a(n+1)-an = 2n+1
an-a(n-1) = 2n-1
an - a2= 5+7+...+(2n-1) (1)
an - a1= 3+5+7+...+(2n-1) (2)
(1)+(2)
2an - (a1+a2) = (n+2)(n-2) + (n+1)(n-1)
= 2n^2-5
2an = 2n^2
an = n^2
(2)
1/an = 1/n^2
for n>=2
1/an = 1/n^2
< 1/[n(n-1)]
= 1/(n-1) - 1/n
Tn = 1/a1+1/a2+.+1/an
= 1+1/a2+.+1/an