(1)∵an+an+1=2n,∴an+1-·2n+1=-, =-1,∴是等比数列, 又a1-=,q=-1,∴an= [2n-(-1)n]. (2)由(1)得Sn=a1+a2+…+an = (2+22+…+2n)- [(-1)+(-1)2+…+(-1)n]= = (3)∵bn=an·an+1, ∴bn=[2n-(-1)n][2n+1-(-1)n+1]=[22n+1-(-2)n-1],∴bn-t·Sn>0, ∴[22n+1-(-2)n-1]-t·>0,∴当n为奇数时, (22n+1+2n-1)-(2n+1-1)>0,∴t< (2n+1)对任意的n为奇数都成立,∴t<1. ∴当n为偶数时, (22n+1-2n-1)-(2n+1-2)>0, ∴ (22n+1-2n-1)- (2n-1)>0, ∴t< (2n+1+1)对任意的n为偶数都成立,∴t<. 综上所述,t的取值范围为t<1 |