计算1/2+(1/4+2/4+3/4)+(1/6+/2/6+3/6+4/6+5/6)+``````+(1/2010+2/2012+`````2009/2010)
问题描述:
计算1/2+(1/4+2/4+3/4)+(1/6+/2/6+3/6+4/6+5/6)+``````+(1/2010+2/2012+`````2009/2010)
答
通项an=(1+2+3+……+2n-1)/(2n)=[(2n-1)(1+2n-1)/2]/(2n)=(2n-1)/2=n-(1/2)a1=1/2,a(n+1)-an=1{an}是首项为1/2,公差为1的等差数列所求的式子是{an}的前1005项的和记为S1005=1005*(1/2+1005-1/2)/2=1005*1005/2=50501...