数学归纳法证明数学归纳法：1-1/2+
1,当n=1时,左边=1-1/2=1/2
右边=1/2,左边等于右边,所以此时等式成立
2,假设n=k(k>=2,k为整数)时等式也成立,则有
1-1/2+1/3-1/4+……1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+……1/2k
当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/(2k+1)+1/(k+1)-1/2(k+1)
=1/(k+2)+……1/2k+1/(2k+1)+1/2...全部
1,当n=1时,左边=1-1/2=1/2
右边=1/2,左边等于右边,所以此时等式成立
2,假设n=k(k>=2,k为整数)时等式也成立,则有
1-1/2+1/3-1/4+……1/(2k-1)-1/2k=1/(k+1)+1/(k+2)+……1/2k
当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/(2k+1)+1/(k+1)-1/2(k+1)
=1/(k+2)+……1/2k+1/(2k+1)+1/2(k+1)=右边
所以此时等式也成立
综合1,2可得证原命题成立。
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