设数列满足a1+3a1+32a1+…+3n-1an=.
(Ⅰ)求数列的通项;
(Ⅱ)设bn=,求数列的前n项和Sn.
(I) ∵ ①
∴当时, ②
①-②得
∴
验证时也满足上式,
(Ⅱ) ∵
∴,
∴ ③
④
③-④得
,