设等差数列{a n }的前n项和为S n ,公差d>0,若a 2 =2,a 5 =11.(1)求数列{a n }的通项公式;(2)设

2024-12-19 15:15:40
推荐回答(1个)
回答1:

(1)设等差数列{a n }的通项为a n =a 1 +(n-1)d,
由题得:a 1 +d=2,a 1 +4d=11,(2分)
解得:a 1 =-1,d=3,a n =3n-4(4分)
(2)由(1)得: S n =
n(3n-5)
2
(6分)
b n =
n(3n-5)
2(n+a)

b 1 =
-1
1+a
b 2 =
1
2+a
b 3 =
6
3+a

∵{b n }是等差数列,
2
2+a
=
-1
1+a
+
6
3+a

a=-
5
3
b n =
3n
2
(8分)
又∵ c n = 2 b 2n = 2 3n
c 1 + c 2 +…+ c n =
8
7
( 8 n -1)
(10分)
lim
n→+∞
c 1 + c 2 +…+ c n
b n +1
=
8
7
( 8 n -1)
b n +1
=
0(|b<8)
8
7
(b=8)
不存在(|b<8或b=-8)
(12分)