(2014•攀枝花二模)下列离子方程式中,正确的是(  )

1个回答

  • (1)证明:由已知,4Sn=

    a2n+2an,且an>0.

    当n=1时,4a1=

    a21+2a1,解得a1=2.

    当n≥2时,有4Sn−1=

    a2n−1+2an−1.

    于是4Sn−4Sn−1=

    a2n−

    a2n−1+2an−2an−1,即4an=

    a2n−

    a2n−1+2an−2an−1.

    于是

    a2n−

    a2n−1=2an+2an−1,即(an+an-1)(an-an-1)=2(an+an-1).

    ∵an+an-1>0,

    ∴an-an-1=2(n≥2).

    故数列{an}是首项为2,公差为2的等差数列,且an=2n;

    (2)证明:∵an=2n,则[1

    Sn=

    1

    n(n+1)=

    1/n−

    1

    n+1],

    ∴[1

    S1+

    1

    S2+…+

    1

    Sn=(1−

    1/2)+(

    1

    2−

    1

    3)+…+(

    1

    n−

    1

    n+1)=1−

    1

    n+1<1.

    ∵1−

    1

    n+1]随着n的增大而增大,

    ∴当n=1时取最小值[1/2].

    故原不等式成立;

    (3)由