1-1/2+1/3-1/4+……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n 用数学归纳法证明证明步骤略掉了,关键就是缺如何化简啊。。。
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![1-1/2+1/3-1/4+……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n 用数学归纳法证明证明步骤略掉了,关键就是缺如何化简啊。。。](/uploads/image/z/941216-32-6.jpg?t=1-1%2F2%2B1%2F3-1%2F4%2B%E2%80%A6%E2%80%A6%2B1%2F%EF%BC%882n-1%EF%BC%89-1%2F2n%3D1%2F%28n%2B1%29%2B1%2F%28n%2B2%29%2B%E2%80%A6%E2%80%A61%2F2n+%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E%E8%AF%81%E6%98%8E%E6%AD%A5%E9%AA%A4%E7%95%A5%E6%8E%89%E4%BA%86%EF%BC%8C%E5%85%B3%E9%94%AE%E5%B0%B1%E6%98%AF%E7%BC%BA%E5%A6%82%E4%BD%95%E5%8C%96%E7%AE%80%E5%95%8A%E3%80%82%E3%80%82%E3%80%82)
1-1/2+1/3-1/4+……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n 用数学归纳法证明证明步骤略掉了,关键就是缺如何化简啊。。。
1-1/2+1/3-1/4+……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n 用数学归纳法证明
证明步骤略掉了,关键就是缺如何化简啊。。。
1-1/2+1/3-1/4+……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n 用数学归纳法证明证明步骤略掉了,关键就是缺如何化简啊。。。
当n=1时,1-1/2=1/2成立
假设n=k(k≥1)时等式成立,即:
1-1-1/2+1/3-1/4+……+1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+……1/2k
则n=k+1时
左边=(1-1/2+1/3-1/4+……+1/(2k-1)-1/2k)+(1/2k+1)-(1/2k+2)
=(1/(k+1)+1/(k+2)+……1/2k)+(1/2k+1)-(1/2k+2)
=1/(k+2)+……1/2k+【1/(k+1)+(1/2k+1)-(1/2k+2)】
=1/(k+2)+……1/2k+【(1/2k+1)+(1/2k+2)】
所以当n=k+1时等式也成立
综上,等式成立,得证
n=1
1-1/2=1/2
成立
假设n=k成立,k≥1
1-1/2+1/3-1/4+……+1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+……+1/2k
则n=k+1
1-1/2+1/3-1/4+……+1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2)
=1/(k+1)+1/(k+2)+……+1/2k+1/(2k+1...
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n=1
1-1/2=1/2
成立
假设n=k成立,k≥1
1-1/2+1/3-1/4+……+1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+……+1/2k
则n=k+1
1-1/2+1/3-1/4+……+1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2)
=1/(k+1)+1/(k+2)+……+1/2k+1/(2k+1)-1/(2k+2)
=1/(k+2)+……+1/2k+1/(2k+1)+[1/(k+1)-1/(2k+2)]
=1/[(k+1)+1]+……+1/2k+1/(2k+1)+[2/(2k+2)-1/(2k+2)]
=1/[(k+1)+1]+……+1/[2(k+1)]
成立
综上
原命题成立
收起