分析:(Ⅰ)利用等比數(shù)列通項(xiàng)公式得,數(shù)列{2an-1}的通項(xiàng)公式,移向變形得出數(shù)列{an}的通項(xiàng)公式;
(Ⅱ)由Sn=2n2+2n-2,利用當(dāng)n=1時(shí),b1=S1,當(dāng)n≥2時(shí),bn=Sn=-S n-1,求出bn,繼而求出cn,選擇適當(dāng)求和法求解.
解答:解:(Ⅰ)根據(jù)等比數(shù)列通項(xiàng)公式得,數(shù)列{2a
n-1}的通項(xiàng)公式為2a
n-1=1×3
n-1,所以a
n=
(Ⅱ)由S
n=2n
2+2n-2,可得
當(dāng)n=1時(shí),b
1=S
1=2,
當(dāng)n≥2時(shí),b
n=S
n=-S
n-1=4n,
所以c
n=(a
n-
)•b
n=
•b
n=
當(dāng)n=1時(shí),T
1=c
1=1,
當(dāng)n≥2時(shí),T
n=1+4•3+6•3
2+…+2n•3
n-13T
n=3+4•3
2+6•3
3+…+2(n-1)•3
n-1+2n•3
n-2T
n=-2+4×3+2•3
2+2•3
3+…+2•3
n-1-2n•3
n=10+
-2n•3
n=(1-2n)•3
n+1,
T
n=
.
n=1時(shí),也適合,所以T
n=
.
點(diǎn)評:本題考查數(shù)列通項(xiàng)公式求解,數(shù)列求和,本題中要注意分類討論求通項(xiàng)和求和,是本題的易錯(cuò)點(diǎn).屬于中檔題.