已知等比数列{a n }的各项都是正数,前n项和为S n ,且a 3 =4,S 4 =s 2 +12,求:(1)首项a 1 及公比q

2025-04-06 18:12:08
推荐回答(1个)
回答1:

(1)由S 4 -S 2 =12,得a 3 +a 4 =12,则a 4 =8
q=
a 4
a 3
=
8
4
=2, a 1 =
a 3
q 2
=1
(5分)
(2)由(1)知:数列{a n }的首项为1,公比为2的等比数列,
a n =2 n-1 ,b n =n?2 n-1
T n = b 1 + b 2 + b 3 + b n =1+2?2+3? 2 2 ++n? 2 n-1
2 T n =2+2? 2 2 +3? 2 3 ++(n-1)? 2 n-1 +n? 2 n
T n =(n-1) 2 n +1

故数列数列{b n }的前项和T n 为(n-1)2 n +1(12分)