已知数列{an}的前n项和为Sn,且2Sn=3an-2n(n∈N+)

1个回答

  • ∵2Sn=3an-2n (1)

    ∴2S(n+1)=3a(n+1)-2(n+1) (2)

    (2)-(1):

    2a(n+1)=2S(n+1)-Sn

    =3a(n+1)-2(n+1)-3an+2n

    =3a(n+1)-3an-2

    ∴a(n+1)=3an+2

    ∴[1+a(n+1) ]/(1+an)=3

    ∴ {1+an}是等比数列,公比为3

    ∵2a1=2S1=3a1-2

    ∴a1=2

    ∴1+an=3^n

    ∴an=3^n-1

    2

    ∵ an/[1+a(n+1)]=(3^n-1)/[1+3^(n+1)-1]

    =(3^n-1)/3^(n+1)=1/3-1/3^(n+1)

    ∴Tn=a1/(1+a2)+a2/(1+a3)+……+an/(1+a(n+1)),

    =(1/3-1/3^2)+(1/3-1/3^3)+.+[1/3-1/3^(n+1)]

    =n/3-[1/3^2+1/3^3+.+1/3^(n+1)]

    =n/3-1/9(1-/3^n)/(1-1/3)

    =n/3-1/6+1/6*1/3^n (∵1/6*1/3^n>0)

    >n/3-1/6=(2n-1)/6