(Ⅰ)由ni=1=Sn2, (1) 由n+1i=1=Sn+12, (2) (2)-(1),得=(Sn+1+Sn)(Sn+1-Sn)=(2 Sn+an+1) an+1. ∵ an+1 >0,∴an+12-=2Sn. 由an+12-=2Sn,及an2-an =2Sn-1 (n≥2), 两式相减,得(an+1+ an)( an+1-an)= an+1+ an. ∵an+1+ an >0,∴an+1-an =1(n≥2) 当n=1,2时,易得a1=1,a2=2,∴an+1- an =1(n≥1). ∴{ an}成等差数列,首项a1=1,公差d=1,故an=n. (Ⅱ)由,得。所以, 当时,; 当时, ,
即 (Ⅲ)nk=1=nk=1<1+nk=2 <1+nk=2= =1+nk=2 (-) =1+1+--<2+<3. |