在各项都为正数的等比数列{a n }中,已知a 3 =4,前三项的和为28.
1个回答
展开全部
(Ⅰ)设公比为q,则有a 3 =4,前三项的和为28,
知
a 1 q 2 =4
a 1 (1- q 3 )
1-q =28 ,
解得 a 1 =16,q=
1
2 ,或 a 1 =36,q=-
1
3 .
∵等比数列{a n }各项都为正数,
∴ a 1 =36,q=-
1
3 不合题意,舍去.
∴ a 1 =16,q=
1
2 ,
a n =16× (
1
2 ) n-1 =32× (
1
2 ) n .
(Ⅱ)∵ a n =32× (
1
2 ) n ,
∴b n =log 2 a n = log 2 [32×(
1
2 ) n ] =5-n.
S n =b 1 +b 2 +…+b n =4+3+2+…+(5-n)
=
n(9-n)
2 .
∴
S n
n =
9-n
2 ,
∴
S 1
1 +
S 2
2 +…+
S n
n =
9-1
2 +
9-2
2 +…+
9-n
2
=
9n
2 -
n(n+1)
2
=-(
1
2 n 2 -4n )
= -
1
2 (n-4 ) 2 +8 .
∴n=4时,
S 1
1 +
S 2
2 +…+
S n
n 取最大值8.
知
a 1 q 2 =4
a 1 (1- q 3 )
1-q =28 ,
解得 a 1 =16,q=
1
2 ,或 a 1 =36,q=-
1
3 .
∵等比数列{a n }各项都为正数,
∴ a 1 =36,q=-
1
3 不合题意,舍去.
∴ a 1 =16,q=
1
2 ,
a n =16× (
1
2 ) n-1 =32× (
1
2 ) n .
(Ⅱ)∵ a n =32× (
1
2 ) n ,
∴b n =log 2 a n = log 2 [32×(
1
2 ) n ] =5-n.
S n =b 1 +b 2 +…+b n =4+3+2+…+(5-n)
=
n(9-n)
2 .
∴
S n
n =
9-n
2 ,
∴
S 1
1 +
S 2
2 +…+
S n
n =
9-1
2 +
9-2
2 +…+
9-n
2
=
9n
2 -
n(n+1)
2
=-(
1
2 n 2 -4n )
= -
1
2 (n-4 ) 2 +8 .
∴n=4时,
S 1
1 +
S 2
2 +…+
S n
n 取最大值8.
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询