作业帮对任意的正偶数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]](/uploads/image/z/3746039-23-9.jpg?t=%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84%E6%AD%A3%E5%81%B6%E6%95%B0n%2C%E6%B1%82%E8%AF%811-1%2F2%2B1%2F3.%2B1%2F%28n-1%29-1%2Fn%3D2%5B1%2F%28n%2B2%29%2B%281%2Fn%2B4%29%2B.%2B1%2F2n%5D)
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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]
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]
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