(1)在2Sn=an+1﹣2n+1+1中, 令n=1得:2S1=a2﹣22+1, 令n=2得:2S2=a3﹣23+1, 解得:a2=2a1+3,a3=6a1+13 又2(a2+5)=a1+a3 解得a1=1 (2)由2Sn=an+1﹣2n+1+1, 得an+2=3an+1+2n+1, 又a1=1,a2=5也满足a2=3a1+21, 所以an+1=3an+2n对n∈N*成立 ∴an+1+2n+1=3(an+2n),又a1=1,a1+21=3, ∴an+2n=3n, ∴an=3n﹣2n; (3)(法一) ∵an=3n﹣2n=(3﹣2)(3n﹣1+3n﹣2×2+3n﹣3×22+…+2n﹣1)≥3n﹣1 ∴≤, ∴+++…+≤1+++…+=<; (法二)∵an+1=3n+1﹣2n+1>2×3n﹣2n+1=2an, ∴<•,, 当n≥2时,<•,<•,, …<•, 累乘得:<•, ∴+++…+≤1++×+…+×<<. |