(1)在数列{an}中,∵an+1=an+1,∴an+1-an=1 则数列{an}是公差为1的等差数列,又a1=6, ∴an=a1+(n-1)d=6+1×(n-1)=n+5. 设l上任意一点P(x,y),∵点A(0,1)在直线l上,则=(x,y-1), 由已知可得∥,又向量=(1,2), ∴2x-(y-1)=0,∴直线l的方程为y=2x+1, 又直线l过点(n,bn),∴bn=2n+1; (2)由cn=n•2bn=n•22n+1 ∴Sn=C1+C2+…+cn =1×23+2×25+3×27+…+n•22n+1① 4Sn=1×25+2×27+…+(n-1)•22n+1+n•22n+3② ①-②得:-3Sn=23+25+27+…+22n+1-n•22n+3. =-n•22n+3=-n•22n+3 ∴Sn=. |