(1)由题设条件知bn+1-bn=2n.由此能够求出数列{bn}的通项公式.
(2)做差比较,由bn•bn+2-bn+12=(2n-1)•(2n+2-1)-(2n+1-1)2=(22n+2-2n+2-2n+1)-(22n+2-2n+2+1)=-2n,与0比较可得答案.
【解析】
(1)由已知得an=n.从而bn+1=bn+2n,即bn+1-bn=2n.(2分)
∴bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1=.(6分)
(2)因为bn•bn+2-bn+12=(2n-1)•(2n+2-1)-(2n+1-1)2
=(22n+2-2n+2-2n+1)-(22n+2-2n+2+1)=-2n<0,
∴bn•bn+2<bn+12.(12分)