1/(1×3)+1/+……+1/[(2n-1)×(2n+1)]=n/(2n+1)
证:
当n=1时,1/(1×3)=1/(2×1-1),命题成立;
设当n=k时,命题成立,
即:1/(1×3)+1/+……+1/[(2k-1)×(2k+1)]=k/(2k+1);
当n=k+1时,
1/(1×3)+1/+……+1/[(2k-1)×(2k+1)]+1/[(2k+1)(2k+3)]
=k/(2k+1)+1/[(2k+1)(2k+3)]
=k(2k+3)/[(2k+1)(2k+3)]+1/[(2k+1)(2k+3)]
=(2k^2+3k+1)/[(2k+1)(2k+3)]
=(2k+1)(k+1)/[(2k+1)(2k+3)]
=(k+1)/(2k+3)
=(k+1)/[2(k+1)+1]
即:
1/(1×3)+1/+……+1/[(2k-1)×(2k+1)]+1/[(2k+1)(2k+3)]=(k+1)/[2(k+1)+1]
命题成立.
所以:1/(1×3)+1/+……+1/[(2n-1)×(2n+1)]=n/(2n+1)
证毕.