m=1 mn=2 1/mn+1/(n+1)(m+1)+1/(n+2)(m+2)+1/(n+3)(m+3)...+1/(n+2010)(m+2010)=多少?

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/14 22:36:05

m=1 mn=2 1/mn+1/(n+1)(m+1)+1/(n+2)(m+2)+1/(n+3)(m+3)...+1/(n+2010)(m+2010)=多少?
m=1 mn=2
1/mn+1/(n+1)(m+1)+1/(n+2)(m+2)+1/(n+3)(m+3)...+1/(n+2010)(m+2010)
=多少?

m=1 mn=2 1/mn+1/(n+1)(m+1)+1/(n+2)(m+2)+1/(n+3)(m+3)...+1/(n+2010)(m+2010)=多少?
因为:m=1 mn=2,所以:m=1 n=2/m=2,
把m=1,n=2代入下面式子中,便可得解.
即为:1/1*2+1/(2*3)+1/(3*4)...+1/(2011*2012)
发现没有1/2=1/1-1/2
1/(2*3)=1/2-1/3
1/(3*4)=1/3-1/4依此类推,
1/(2011*2012)=1/2011-1/2022
所以原式=1/2+1/2-1/3+1/3-1/4+1/4-1/5...+1/2010-1/2011+1/2011-1/2012,中间是抵销了,=1-1/2012
=2011/2012

(1-1/2)+(1/2-1/3)+1/3-1/4)