(1)由S9=63,解得a5=7.由a1+a5=6,得a1=-1,故d=,由此能求出数列{an}的通项公式an及前n项和Sn.
(2)由,知an•bn=(2n-3)•22n-3,故+…+(2n-3)•22n-3,利用错位相减法能求出数列{an•bn}的前n项和Tn.
【解析】
(1)∵S9=63,∴9a5=63,解得a5=7.
∵a1+a5=6,∴a1=-1,
∴d=,
∴an=2n-3,.
(2)∵an=2n-3,,
∴,
∴an•bn=(2n-3)•22n-3,
+…+(2n-3)•22n-3,
4Tn=-1×21+1•23+3•25+…+(2n-5)•22n-3+(2n-3)•22n-1,
两式相减,得:-3Tn=-
=-
=,
.