vbtyuq
幼苗
共回答了23个问题采纳率:78.3% 举报
(本小题满分14分)
(Ⅰ)设等比数列{a n }的公比为q,∵a n >0,∴q>0
若q=1时S m =ma 1 S 2m =2ma 1 ,此时2S m =S 2m ,而已知S m =26,S 2m =728,∴2S m ≠S 2m ,∴q=1不成立…(1分)
若q≠1,由
S m =26
S m =728 得
a 1 (1- q m )
1-q =26(1)
a 1 (1- q 2m )
1-q =728(2) …(2分)
(1)÷(2)得:1+q m =28∴q m =27…(3分)
∵q m =27>1∴q>1
∴前m项中a m 最大∴a m =18…(4分)
由 a 1 q m-1 =18 得,
a 1 q m-1
q m =
18
27 ∴
a 1
q =
2
3 (3) 即 a 1 =
2
3 q
把 a 1 =
2
3 q 及q m =27代入(1)式得
2
3 q(1-27)
1-q =26
解得q=3
把q=3代入 a 1 =
2
3 q 得a 1 =2,所以 a n =2× 3 n-1 …(7分)
由 T n =2 n 2
(1)当n=1时 b 1 =T 1 =2
(2)当 n≥2时 b n = T n - T n-1 =2 n 2 -2(n-1 ) 2 =2 n 2 -2( n 2 -2n+1) =4n-2
∵b 1 =2适合上式∴b n =4n-2…(9分)
(Ⅱ)由(1)得 c n =(4n-2)•2× 3 n-1 =4(2n-1)× 3 n-1
记 d n =(2n-1)× 3 n-1 ,d n 的前n项和为Q n ,显然P n =4Q n Q n = d 1 + d 2 + d 3 +…+ d n =1× 3 0 +3× 3 1 +5× 3 2 +…+(2n-1)× 3 n-1 …①∴ 3 Q n = d 1 + d 2 + d 3 +…+ d n =1× 3 1 +3× 3 2 +5× 3 3 +…+(2n-1)× 3 n …..②
…(11分)
①-②得:-2Q n =1+2×3 1 +2×3 2 +2×3 3 +…2×3 n-1 -(2n-1)×3 n
= 1+2×
3(1- 3 n-1 )
1-3 -(2n-1)× 3 n =-2-(2n-2)×3 n …(13分)
∴ 4 Q n =4(n-1)× 3 n +4 ,
即 P n =4(n-1)× 3 n +4 …(14分)
1年前
7