(Ⅰ)由数列{}是等比数列,设该数列的公比为q,则=•q3,又a2=18,a5=1215,可求得q=3,从而可求得an.
(Ⅱ)设Sn=a1+a2+a3+…+an,则Sn=3+2×32+3×33+…+n•3n,利用错位相减法即可求得Sn的值.
解(I):由题意得:数列{}是等比数列,设该数列的公比为q,
则=•q3…2′
又a2=18,a5=1215,
∴q3=27,q=3…4′
∴=•qn-2,
∴an=n•3n…6′
(Ⅱ)设Sn=a1+a2+a3+…+an,
则Sn=3+2×32+3×33+…+n•3n,①
∴3Sn=32+2×33+…+(n-1)•3n+n•3n+1②…8′
①-②
-2Sn=3+32+33+…+3n-n•3n+1…10′
=-n•3n+1
=•3n+1-,
∴Sn=…12′