已知首项为 3 2 的等比数列{a n }的前n项和为S n (n∈N * ),且-2S 2 ,S 3 ,4S 4 成等差

2025-04-07 21:55:16
推荐回答(1个)
回答1:

(Ⅰ)设等比数列{a n }的公比为q,
∵-2S 2 ,S 3 ,4S 4 等差数列,
∴2S 3 =-2S 2 +4S 4 ,即S 4 -S 3 =S 2 -S 4
得2a 4 =-a 3 ,∴q= -
1
2

a 1 =
3
2
,∴ a n =
3
2
? (-
1
2
)
n-1
= (-1) n-1 ?
3
2 n

(Ⅱ)证明:由(Ⅰ)得,S n =
3
2
[1-(-
1
2
) n ]
1+
1
2
=1- (-
1
2
)
n

S n +
1
S n
=1 -(-
1
2
)
n
+
1
1 -(-
1
2
)
n

当n为奇数时, S n +
1
S n
=1 +(
1
2
)
n
+
1
1 +(
1
2
)
n
= 1+
1
2 n
+
2 n
1 +2 n
= 2+
1
2 n (2 n +1)

当n为偶数时, S n +
1
S n
=1 -(
1
2
)
n
+
1
1 -(
1
2
)
n
= 2+
1
2 n (2 n -1)

S n +
1
S n
随着n的增大而减小,
S n +
1
S n
S 1 +
1
S 1
=
13
6
,且 S n +
1
S n
S 2 +
1
S 2
=
25
12

综上,有 S n +
1
S n
13
6
(n∈N*)
成立.