对任意的正偶数n,求证1-1/2+1/3.+1/(n-1)-1/n=2[1/(n+2)+(1/n+4)+.+1/2n]
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对任意的正偶数n,求证1-1/2+1/3.+1/(n-1)-1/n=2[1/(n+2)+(1/n+4)+.+1/2n]
证明:
当n=2时,左边=1-1/2=1/2,右边=2(1/(2+2))=1/2,左边=右边,成立
假设当n=2k时,成立,即1-1/2+1/3.+1/(2k-1)-1/2k=2[1/(2k+2)+(1/2k+4)+.+1/(4k)]
则当n=2k+2时,左边=1-1/2+1/3.+1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2)
因为1-1/2+1/3.+1/(2k-1)-1/2k=2[1/(2k+2)+1/(2k+4)+.+1/(4k)]
所以左边=2[1/(2k+2)+1/(2k+4)+...+1/(4k)]+1/(2k+1)-1/(2k+2)=2[1/(2k+2)+1/(2k+4)+...+1/(4k)]+2/(4k+2)-2/(4k+4)
=2[2/(4k+4)+1/(2k+4)+...+1/(4k)+1/(4k+2)-1/(4k+4)]
=2[1/(2k+4)+1/(4k+6)+...+1/(4k)+1/(4k+2)+1/(4k+4)]
=右边
所以成立
证毕
当n=2时,左边=1-1/2=1/2,右边=2(1/(2+2))=1/2,左边=右边,成立
假设当n=2k时,成立,即1-1/2+1/3.+1/(2k-1)-1/2k=2[1/(2k+2)+(1/2k+4)+.+1/(4k)]
则当n=2k+2时,左边=1-1/2+1/3.+1/(2k-1)-1/2k+1/(2k+1)-1/(2k+2)
因为1-1/2+1/3.+1/(2k-1)-1/2k=2[1/(2k+2)+1/(2k+4)+.+1/(4k)]
所以左边=2[1/(2k+2)+1/(2k+4)+...+1/(4k)]+1/(2k+1)-1/(2k+2)=2[1/(2k+2)+1/(2k+4)+...+1/(4k)]+2/(4k+2)-2/(4k+4)
=2[2/(4k+4)+1/(2k+4)+...+1/(4k)+1/(4k+2)-1/(4k+4)]
=2[1/(2k+4)+1/(4k+6)+...+1/(4k)+1/(4k+2)+1/(4k+4)]
=右边
所以成立
证毕
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