求证1-1/2+1/3-1/4……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n
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求证1-1/2+1/3-1/4……+1/(2n-1)-1/2n=1/(n+1)+1/(n+2)+……1/2n
n为正整数
n为正整数
可以用数学归纳法证明:
如下:
当n=1时,左侧=1-1/2=1/2,右侧=1/2,结论成立;
假设n=k成立,则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,左侧={1-1/2+1/3-1/4……+1/(2k-1)-1/2k}+1/(2k+1)-1/(2k +2)
右侧=1/(k+2)+……1/2k+1/(2k+1)+1/(2k +2)={1/(k+1)+1/(k+2)+……1/2k}+1/(2k+1)+1/(2k +2)-1/(k+1)=)={1/(k+1)+1/(k+2)+……1/2k}+1/(2k+1)-1/(2k +2)
根据假设,所以当n=k+1时,左侧=右侧,
所以.
如下:
当n=1时,左侧=1-1/2=1/2,右侧=1/2,结论成立;
假设n=k成立,则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,左侧={1-1/2+1/3-1/4……+1/(2k-1)-1/2k}+1/(2k+1)-1/(2k +2)
右侧=1/(k+2)+……1/2k+1/(2k+1)+1/(2k +2)={1/(k+1)+1/(k+2)+……1/2k}+1/(2k+1)+1/(2k +2)-1/(k+1)=)={1/(k+1)+1/(k+2)+……1/2k}+1/(2k+1)-1/(2k +2)
根据假设,所以当n=k+1时,左侧=右侧,
所以.
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