(1)求使（an+a n+1）(an+2+an+3)+(an+2-an)＞(an+1+an+2)2-1(n∈N*)成立的d的取值范围；
(2)若bn=a2n-1+a2n(n∈N*),求bn的表达式.
(1)求使（an+a n+1）(an+2+an+3)+(an+2-an)＞(an+1+an+2)2-1(n∈N*)成立的d的取值范围；
(2)若bn=a2n-1+a2n(n∈N*),求bn的表达式.
解：(1)∵a1+a2=3,
∴an+an+1=(a1+a2)+(n-1)d=3+(n-1)d.
∴an+1+an+2=3+nd,an+2+an+3=3+(n+1)d.
又an+2-an=(an+1+an+2)-(an+1+an)=d,
(an+an+1)(an+2+an+3)+(an+2-an)＞(an+1+an+2)2-1,
∴［3+(n-1)d］［3+(n+1)d］+d＞(3+nd)2-1.
∴d2-d-1＜1.
解得0＜d＜
(2)b1=a1+a2=3.
又bn=a2n-1+a2n，bn+1=a2n+1+a2n+2,
∴bn+1-bn=（a2n+1+a2n+2）-(a2n-1+a2n)
=(a2n+1-a2n-1)+(a2n+2-a2n).
∵a2n+1-a2n-1=(a2n+a2n+1)-(a2n-1+a2n)=d,a2n+2-a2n=d,
∴bn+1-bn=2d.
∴{bn}是以首项b1=3,公差为2d的等差数列.
∴bn=3+2（n-1）d=2dn+3-2d(n∈N*).