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.