(1)用数学归纳法证明1+4+7+…+(3n−2)=12n(3n−1)

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  • 解题思路:(1)利用数学归纳法的证题步骤分当n=1时证明等式成立,假设当n=k时等式成立,去证明当n=k+1时等式也成立即可;

    (2)当n=1时,证明左边[1/1×3]与右边[1/2×1+1]相等,假设n=k时等式成立,去证明当n=k+1时等式也成立即可.

    证明:(1)①当n=1时,左边=3×1-2=1,右边[1/2]×1(3×1-1)=1,左边=右边,等式成立;

    ②假设当n=k时等式成立,即1+4+7+…+(3k-2)=[1/2]k(3k-1),

    则当n=k+1时,

    1+4+7+…+(3k-2)+[3(k+1)-2]

    =[1/2]k(3k-1)+[3(k+1)-2]

    =[1/2](3k2+5k+2)

    =[1/2](k+1)(3k+2)

    =[1/2](k+1)[3(k+1)-1],

    即n=k+1时,等式也成立;

    综合①②知,对任意n∈N*,等式成立.

    (2)证明:①当n=1时,证明左边=[1/1×3],右边=[1/2×1+1],左边=右边,等式成立;

    ②假设当n=k时等式成立,即[1/1×3]+[1/3×5]+…+[1

    (2k−1)(2k+1)=

    k/2k+1],

    则当n=k+1时,

    [1/1×3]+[1/3×5]+…+[1

    (2k−1)(2k+1)+

    1

    [2(k+1)−1][2(k+1)+1]

    =

    k/2k+1]+[1

    [2(k+1)−1][2(k+1)+1]

    =

    k(2k+3)+1

    [2(k+1)−1][2(k+1)+1]

    =

    (k+1)(2k+1)

    [2(k+1)−1][2(k+1)+1]

    =

    k+1/2k+3]

    =[k+1

    [2(k+1)+1],

    即当n=k+1时,等式也成立;

    综上知,对任意n∈N*,等式

    1/1×3]+[1/3×5]+…+[1

    (2n−1)(2n+1)=

    n/2n+1]恒成立.

    点评:

    本题考点: 数学归纳法.

    考点点评: 本题考查数学归纳法,着重考查利用数学归纳法证明问题,考查推理与逻辑思维能力,属于中档题.