Question

已知数列an中,a1=1,an+1=an+2n+1,求数列an的通项公式。（重点是后面的！！！谢啦）

令bn=(2n+1)/an*an+1,数列bn的前n项和为Tn,Tn>m恒成立，求m的取值范围

Answer 1

a(n+1) = an +2n+1
a(n+1) -an = 2n+1
an -a(n-1) = 2n-1
an -a1 = [3+5+....+(2n-1)]
an = 1+3+5+....+(2n-1)
=n^2

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

bn = (2n+1)/[n^2.(n+1)^2] > 0
Tn = b1+b2+....+bn
≥ T1
=3/4
Tn > m
m < 3/4

追问

为啥我第一问算的是n^2+2n+1？？？我是用an=an-1+2(n-1)+1,an-1=an-2+2(n-2)+1……，a2=a1+2+1,然后所有等式相加的累加法、、嘿嘿、大神、谢啦

追答

其实我也是用累加法

a(n+1) = an +2n+1
a(n+1) -an = 2n+1

an -a(n-1) = 2n-1 (1)
a(n-1) -a(n-2) = 2n-3 (2)
....
....
....
a2-a1 = 3 (n-1)
累加法

(1)+(2)+(3)+...+(n-1)
an - a1 = 3+5+7+.....+2n-1
an = 1+3+5+7+...+2n-1
= n^2