设M=1/(1+√2)+1/(√2+√3)+1/(√3+√4)+…+1/(√2011+√2012),N=1-2+3-4+5-6+…+2011-2012,求

问题描述:

设M=1/(1+√2)+1/(√2+√3)+1/(√3+√4)+…+1/(√2011+√2012),N=1-2+3-4+5-6+…+2011-2012,求
求N/(M+1)²

N=1-2+3-4+5-6+…+2011-2012=-1006
M+1=1+1/(1+√2)+1/(√2+√3)+1/(√3+√4)+…+1/(√2011+√2012),
利用平方差公司从第二项处理成:1/(1+√2)=(1-√2)/1-2=-(1-√2)、1/(√2+√3)=(√2-√3)/2-3到最后
M+1=1-(1-√2)-(√2-√3)-(√3-√4)-…-(√2011-√2012)=√2012
N/(M+1)²=1006/(√2012)^2=1/2