摘要:又T1=1.T2=4.∴a1=1.q=2.知:a=1.q=2.∴an=a1?qn-1=2n-1∴Tn=n?1+(n-1)?2+(n-2)?22+-+2?2n-2+1?2n-1 ①2Tn=n?2+(n-1)?22+(n-2)?23+-+2?2n-1+1?2n ②②-①得Tn=n?2+(n-1)?22+-+2?2n-1+2n-[n?1+(n-1)2+-+2?2n-2+1?2n-1]=-n+2+22+-+2n-1+2

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