由题有,Tn=2Sn-n^2 ①
∴Tn-1=2Sn-1-（n-1）^2②
①②相减,∴Sn=2an-2n+1③
Sn-1=2an-1-2(n-1)+1④
③④相减,∴an=2an-1+2
∴(an)+2=2(an-1+2)
∴｛(an)+2｝是以a1+2为首项,2为公比的等比数列,∴
an+2=(a1+2)*2^(n-1)
∴an=(a1+2)*2^(n-1)-2