已知数列{an}的前n项和为Sn,且满足Sn+1=4an+2,(n∈N*),a1=2,

1个回答

  • 解题思路:(1)由Sn+1=4an+2,得an+1=Sn+1-Sn=(4an+2)-(4an-1+2)(n≥2),由此能求出λ=2.

    (2)由已知得bn=an+1-2an=4•2n-1,将an+1-2an=4•2n-1两边同除以2n+1,得

    a

    n+1

    2

    n+1

    a

    n

    2

    n

    =4•2-2=1,由此能求出cn=n.

    (3)由

    c

    n

    a

    n

    2

    n

    =n,得

    a

    n

    =n•

    2

    n

    ,从而dn=

    2

    n

    n

    2

    n+1

    n+1

    ,由此能求出

    T

    n

    =2−

    2

    n+1

    n+1

    (1)由Sn+1=4an+2,得an+1=Sn+1-Sn=(4an+2)-(4an-1+2)(n≥2)

    ∴an+1-2an=2an-4an-1=2(an-2an-1

    故数列{an+1-2an} 是以a2-2a1为首项,2为公比的等比数列,

    ∵bn=an+1-λan,数列{bn}为等比数列,

    ∴λ=2.

    (2)由a1=2,a1+a2=S2=4a1+2,

    ∴a2=8,

    ∴bn=an+1-2an=4•2n-1

    将an+1-2an=4•2n-1两边同除以2n+1

    an+1

    2n+1−

    an

    2n=4•2-2=1,即cn+1-cn=1,

    故{cn}是以c1=

    a1

    2=1为首项,1为公差的等差数列,

    ∴cn=n.

    (3)∵cn=

    an

    2n=n,∴an=n•2n,

    ∴dn=([1

    2log2

    an/n]-[1

    log2

    an+1/n+1])•2n+1

    =([1/2n−

    1

    n+1])•2n+1

    =

    2n

    n−

    2n+1

    n+1,

    ∴Tn=2-

    22

    2+

    22

    2−

    23

    3+…+

    2n

    n−

    2n+1

    n+1

    =2-

    2n+1

    n+1.

    ∴Tn=2−

    2n+1

    点评:

    本题考点: 数列的求和;数列递推式.

    考点点评: 本题考查实数值的求法,考查数列的通项公式的求法,考查数列的前n项和的求法,解题时要认真审题,注意裂项求和法的合理运用.