分析:(1)由數(shù)列的前n項(xiàng)和分類求出a1和an(n≥2),驗(yàn)證a1后可得通項(xiàng)公式;
(2)分別求出數(shù)列{bn}的前3項(xiàng),由等差中項(xiàng)的概念求出c,代入后利用錯(cuò)位相減法求數(shù)列{an•2bn}的前n項(xiàng)和Tn.
解答:解:(1)由S
n=2n
2-n
當(dāng)n=1時(shí),a
1=S
1=1.
當(dāng)n≥2時(shí),
an=Sn-Sn-1=(2n2-n)-[2(n-1)2-(n-1)]=4n-3.
n=1時(shí)成立.
∴a
n=4n-3;
(2)∵數(shù)列{b
n}是等差數(shù)列,
由b
n=
,取n=1得,
b1=.
取n=2得,
b2==.
取n=3得,
b3==.
∴
+=,解得c=
-.
∴b
1=2,公差d=2.
∴b
n=2+2(n-1)=2n.
則a
n•2
bn=(4n-3)•4
n∴
Tn=1×41+5×42+…+(4n-7)×4n-1+(4n-3)×4n①
4Tn=1×42+5×43+…+(4n-7)×4n+(4n-3)×4n+1②
①-②得:
-3Tn=4+4×42+4×43+…+4×4n-(4n-3)×4n+1.
∴
Tn=-(4n-1)+×4n+1.
點(diǎn)評(píng):本題考查了數(shù)列的通項(xiàng)公式,考查了數(shù)列的和,訓(xùn)練了錯(cuò)位相減法,是中檔題.