一道奥数题,简便计算 来自西北工业大学附属中学

数理答疑团为您解答,希望对你有所帮助.
(2m-1)/[(m-1)m(m+1)]
= 2/[(m-1)(m+1)] - 1/[(m-1)m(m+1)]
= 2/[(m-1)(m+1)] - 1/[(m-1)m] * 1/(m+1)
= 2/[(m-1)(m+1)] - [1/(m-1) -1/m] * 1/(m+1)
= 2/[(m-1)(m+1)] - 1/[(m-1)(m+1)] + 1/[m(m+1)]
= 1/[(m-1)(m+1)] + 1/[m(m+1)]
= [1/(m-1) -1/(m+1)]/2 + [1/m - 1/(m+1)]
原式m为3、4、...、10,
故：原式=(1/2 -1/4+1/3-1/5+1/4-1/6+...+1/9-1/11)/2 + (1/3-1/4+1/4-1/5+...+1/10-1/11) + 1
=(1/2 +1/3-1/10-1/11)/2 + (1/3 -1/11) + 1
=1/4 +1/6-1/20-1/22 + 1/3 -1/11 + 1
=86/55
祝你学习进步,更上一层楼! (*^__^*)