饭_团
幼苗
共回答了15个问题采纳率:93.3% 举报
(Ⅰ)∵a 1 =2, S n =
1
2 a n+1 -1 (n∈N * ),
∴当n=1时, S 1 =
1
2 a 2 -1= a 1 =2 ,
解得a 2 =6.
当n=2时, S 2 =
1
2 a 3 -1=2+6=8 ,
解得a 3 =18.
(Ⅱ)∵a 1 =2, S n =
1
2 a n+1 -1 (n∈N * ),
∴当n≥2时, S n =
1
2 a n+1 -1 , S n-1 =
1
2 a n -1 ,
∴ a n = S n - S n-1 =
1
2 a n+1 -
1
2 a n ,
即a n+1 =3a n .
对于a 2 =3a 1 也满足上式,
∴数列{a n }是首项为2,公比为3的等比数列,
∴ a n =2• 3 n-1 (n∈ N * ) .
( III)∵ a n =2• 3 n-1 (n∈ N * ) ,
∴ n a n =2n• 3 n-1 ,
∴ T n =2•1+4•3+6• 3 2 +8• 3 3 +…+2n• 3 n-1 ,
3 T n =2•3+4• 3 2 +6• 3 3 +8• 3 4 +…+2n• 3 n ,
相减得, -2 T n =2(1+3+ 3 2 + 3 3 +…+ 3 n-1 )-2n• 3 n
= 2•
1- 3 n
1-3 -2n• 3 n
=3 n -1-2n•3 n ,
∴ T n =
(2n-1)• 3 n +1
2 .
1年前
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