解:(1)∵an+an+1=2n, ∴an+1﹣ 2n+1=(2n﹣an)﹣ 2n+1 =﹣an+2n(1﹣ )=![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015100212-90876.png) ∴ =﹣1, ∴{an﹣ 2n}是等比数列, 又a1﹣ = ,q=﹣1 ∴an= [2n﹣(﹣1)n]. (2)Sn=a1+a2+…+an= [(2+22+…+2n)﹣((﹣1)+(﹣1)2+…+(﹣1)n)]
![](http://img.shitiku.com.cn/uploads/allimg/20191015/20191015100214-82984.png) (3)∵an,an+1是关于x的方程 的两根, ∴bn=anan+1,bn= [2n﹣(﹣1)n][2n+1﹣(﹣1)n+1] = [2n+1﹣(﹣2)n﹣1] ∵bn﹣msn>0, ∴ , 当n为奇数时, [22n+1+2n﹣1]﹣ (2n+1﹣1)>0, ∴m< (2n+1)对n∈{奇数}都成立, ∴m<1. 当n为偶数时, [22n+1﹣2n﹣1]﹣ (2n+1﹣2)>0, [22n+1﹣2n﹣1]﹣ (2n﹣1)>0, ∴m< (2n+1+1)对n∈{偶数}都成立, ∴m< . 综上所述,m的取值范围为m<1. |