(1)将n=1代入已知递推式,易得a2,从而求出d,故an可求;
(2)求出bn,分p=1和p≠1两种情况讨论,然后利用错位相减法求和.
【解析】
(Ⅰ)设等差数列{an}的公差为d,由得:=3,所以a2=2,即d=a2-a1=1,
所以an=n.
(Ⅱ)由bn=anpan,得bn=npn.所以Tn=p+2p2+3p3+…+(n-1)pn-1+npn,①
当p=1时,Tn=;
当p≠1时,
pTn=p2+2p3+3p4+…+(n-1)pn+npn+1,②
①-②得(1-p)Tn=p+p2+p3+…+pn-1+pn-npn+1=,
即Tn=.